Watch me build an AVL tree step by step with linear data that would normally create a terrible unbalanced BST. See insertions, balance factor calculations, and rotations in action.

Perfect follow-up to my BST and AVL intro videos.

00:00 Introduction to AVL Tree with Bad Data
00:14 Previous Videos on BST and AVL Trees
00:28 Practice Example Building Step by Step
00:50 Adding First Node 12
01:11 Adding Node 21
02:10 Adding Node 30 and First Rotation
04:03 Recomputing Balance Factors
05:08 Adding Node 38
08:03 Adding Node 42 and Second Rotation
12:51 Recomputing Balance After Rotation
14:09 Adding Node 55 and Third Rotation
18:55 Placing Unaccounted Nodes
19:21 Final Tree and Balance Factors
20:07 Conclusion and Thanks

=-=-=-=-=-=-=-=-=

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Hey there! Let’s build an AVL self-balancing binary search tree with some really bad data.

It’s all going to work out, trust me.

Okay, so hopefully by now you’ve seen my other videos where we talked about how to define binary

search trees, how to build them, search through them, add, delete, all that stuff. And also AVL

trees. What are rotations? Why do we rotate? How does the AVL tree maintain itself? And so forth.

just going to be a practice example. In the previous video that I that I

recorded regarding AVL trees, I started off with a really long and gnarly linear

tree and then I suddenly turned on AVLness and we rotated the crap out of

it for about 30 minutes until it was really really good. So now I’m just

going to build a tree one step at a time. Suppose for the sake of argument that

you have some really bad data. You know a long time ago, I love telling this story.

time ago you know my grandma before she passed on she was kind of bitter and she had some enemies in

the neighborhood she used to call the cops on her neighbors for gossip um she had a neighbor who

kind of feuded with her and at one point according to my grandma anyway uh my neighbor like didn’t

like my grandma’s tree and so one day when my grandma was looking out into the backyard

through the back window because she likes to yell at ducks who are swimming in her pool

notice that the other little old lady’s arm just kind of went over my grandma’s fence

with a spray bottle and then sprayed her tree. And then my grandma’s tree died. So this kind of

thing I like to call, you know, tree poisoning or data poisoning or bad data, it can affect a binary

search tree and make it really, really, really slow. But in AVL trees, if we have bad data,

it should actually end up being fine. So I’m just going to say we’re poisoning our grandma’s tree

grandma’s tree with some bad data. I’ll just do maybe maybe five pieces of data. We could go

further than that but it’s it’s going to be a super long video if we do. So we’re going to add

these pieces of data one by one. So that means the first thing I’m going to do remember we are kind

of just regularly you know building a regular binary search tree. It’s just that the differences

after we modify the topology in some way we’ll just check to see that it satisfies the rules of

then we’re going to start performing rotations to make sure that it’s self-balancing.

So we’ll add the 12 first.

So let me see if I can draw a little arrow just indicating,

yeah, we’re going to add this number right now.

I’m going to do an ellipse for a node.

And you know, like for binary, what is that?

Flying saucer for binary search trees.

It’s like, well, if there is no data already, if the tree is empty,

first piece of data you add is going to end up being the root node. So that’s it. We have our

root node here. See my other videos if you want more practice on building trees. So we’re just

going to call that the root node. And then we’re going to compute its balance factor, which is

obviously just going to be a zero. So that was super easy. Now we’re going to move on to adding

the next piece of data. Things are getting a little bit worse. This is going to be a linear tree.

tree if it weren’t for our trusty I guess AVL functionality so the 21 I’m

gonna maybe just duplicate this node right now and set that to a 21 and then

we’ll figure out where does the 21 go well it can’t be the root because the

root is already occupied 21 is larger than 12 so it’s got to be on the right

side so the 21 is going to be the right child of the root node so I’m just going

Connecting line right here.

And then we have to recompute balance factors.

So the 21 is the thing we just added.

We have to compute that.

Then we have to work our way up the tree until we hit root.

So we see the root node.

It now has a balance factor of 1.

So it got a little bit worse.

But we don’t see 2 or worse.

So that means this is currently a valid AVL tree.

And we don’t have to do any rotations.

So now it’s time to add the 30.

just realized I could just upload like a hundred videos like this and I think you’d all be pretty

happy or some people would be pretty happy um I don’t know if I want to do that to myself maybe

I will so we’re going to add the 30 node uh so first I’m just going to duplicate some node

and I’m going to call it 30 then we have to figure out where the 30 node belongs uh it’s not going

to be the root node that’s already occupied 30 belongs on the right side of 12 but we can’t use

of 12 but we can’t use that right child it’s occupied 30 belongs on the right side of 21

so you know it makes sense we called this linear data that would make a linear tree so everything

ended up being on the right side so far it’s just getting worse and worse but now the ablness is

going to kind of come to the rescue we have to recompute the balance factors so the 30 has a 0

the 21 ends up having a 1 again if you want practice computing balance factors see my other

The 12 has a balance factor of 2 and now we know right away this is not a valid ABL tree anymore

because we have a balance factor that is bad. It’s too bad.

So we have to now select an XYZ trinode subtree so we can perform a rotation.

I’m sure you can imagine it’s just going to end up being three nodes that are perfectly balanced like that.

But let’s do it one step at a time.

So the Z node is going to be the 12.

That was the first node that we noticed that was out of whack or like I guess the highest node.

or like I guess the highest node of the lowest nodes.

And then we have to find Y by taking a child of Z.

There’s only one possible child we could choose, so that’s going to be Y.

And then we find X by taking a child of Y.

There’s only one choice again, so it’s just going to be there.

Now we have X, Y, and Z, and I’m going to write it down real fast.

Let me just put this in black ink.

So X is 30, and Y is 21, and Z is 12.

12. Then we’re going to do A and B and C, which are going to be just in order representations of

XYZ. So again, think in your code, you would be doing pointers, you’d probably want to make three

more pointers, call them ABC. If you had a rotation function, the rotation function,

after you selected three nodes to rotate, it would receive XYZ. And then you say, let’s reorder,

you know, XYZ and call them ABC. So A is going to be 12, and B is going to be 21,

it should you know ascend from left to right if we’re not supporting duplicates in our tree

and then i’m going to um just you know make some make some nodes for our target pattern here

our output pattern so i’m going to like make some nodes here kind of place them uh in the perfect

avl tree output pattern a perfectly balanced trinode subtree i’m just going to reconnect this

The output pattern is always the same.

You know, it always starts this way because this is better

than the linear pattern that we have above.

And then I just have to update the numbers.

So the left child is going to be 12.

The right child is going to be 21.

Sorry, the center is going to be 21, the new parent.

And the right child is going to be 30.

Okay, now we look for any nodes that are children of the input nodes

that have been unaccounted for.

You can tell easily that all nodes are being used.

So we don’t even have to worry about that step.

So now that we have our trinode subtree which is perfectly balanced, we are ready to just sort of override it into the diagram.

And since the diagram only has the three nodes to begin with, I could pretty much just like erase everything, call it a day, and say this is the new tree, the new rotated tree.

So I’m just going to maybe put this in the middle I guess.

I’m going to delete this over here and recompute the balance factors.

a zero, get zeros, and this 21 also gets a zero because it’s perfectly balanced.

Next thing is let’s move on to adding the number 38. So we’ll add the 38 here and

same thing that we do before. Let’s make a copy of one of the nodes for our diagram.

Let’s call it the 38 and then let’s figure out where it belongs. It’s not going to be the new

root node because that’s occupied. It belongs on the right side. It’s not going to be the right

It’s not going to be the right child because that’s occupied.

It belongs on the right side of the 30.

It couldn’t go on the left.

That would be an invalid binary search tree.

So I’m just going to stick it right there and do a connection.

And then we’re going to update the balance factors.

So the 38 is a leaf.

It gets 0.

The 30 gets a 1.

And the 21 also gets a 1 because its left subtree is a 1 and its right subtree is 2,

or height, and the difference is just 1.

is just one. Remember we don’t really need to update the 12 or any nodes that are not

directly above the node that we just added or removed. So if you look at this tree there’s

a little bit of imbalance but nothing is two or worse so we don’t actually need to perform a

rotation. This is already a valid AVL tree. So I’m going to move on to adding the next number.

Let’s add a 42. Same thing we’re going to duplicate a node for our diagram and we’re going to call it

going to call it the 42 where does it belong well we start by looking at root that’s occupied 42

belongs on the right side of 21 the right child is occupied 42 belongs on the right side belongs on

the right side i mean it’s always going to be on the very right of the of the diagram because that’s

the poison data that we received uh so i’m going to put it over there and then connect um the child

child relationship and then let’s update the the balance factor so 42 gets a 0

38 gets a 1 and 30 gets a 2 now because it’s got 2 height of subtree on its

right side and no subtree on its left side the 21 we should update left subtree

is a 1 right subtree is a 3 oh it’s still gonna it’s actually gonna be 2 so

we have two instances of the number 2 and again which one do we actually want

by the way here’s here’s how I like to eyeball it you can kind of see oh there’s two levels of

difference in terms of the root node which one do we rotate first well as low as possible because

if we rotate the 30 it might end up fixing the 21 so we don’t want to do any more more work than we

have to so I’m just going to maybe move this over here a little bit and select our xyz um hang on

a second I think I just like didn’t do a snap to grid when I was doing that okay hang on

doing that okay hang on okay so we got to choose our x y and z uh we’re going to say that the 30

is z the lowest node possible that is out of whack and then we hop down to a child of z to find y

we hop down to a child of y to find x and so now we have our xyz i’m just going to write them down

I’m going to change the ink real fast and then we have to get ABC so A is the least value B is the

one that belongs in the middle and C is 42 now we have our ABC we’re ready to draw our output pattern

or target pattern so I’m just going to put like some nodes up here duplicate that a couple times

all right not the prettiest but I’ll take it I’m gonna do my connecting lines

and then I have to update the numbers so a is 30 and then b is 38 and c is 42 still so we’ll just

leave that as is now we have to look for children of any of the output pattern nodes that are

for so we’ll first look at the z node here the 30 it had a right child of 38 that’s already handled

as the y node so we don’t need to worry about the 38 okay next we look at the uh y node it had a

right child of 42 that’s also handled in the output pattern so we don’t have to worry about that uh

then the 42 had no children so we don’t need to really do anything this is going to be super easy

Just as a reminder, the Z could have had a left child, the Y could have had a left child,

and the X could have had two children, at least in this particular pattern.

Oh, I think in my previous video, I didn’t name off all the rotation types.

Whoops.

Okay.

So now I’m ready to just kind of like move this into the diagram.

So the 30 and 38 and 42, we can just kind of get rid of that

move the rotated output pattern into the diagram like this.

We’ll make the 38 the right child of 21.

I want to make it go a little bit more to the right

so that we have a nicely drawn diagram

so that all the nodes have space so it’s easy to debug visually.

It’s easy to just kind of eyeball it.

Then I’m going to update the balance factors.

So the 42 has 0, the 30 has 0, the 38 has 0.

Nice.

The 21, hopefully that improved.

The left subtree has a height of 1, the right subtree has a height of 2.

of one the right subtree has a height of two so actually the balance factor of the 21 node improved

to one we are now done uh rotating and we have a valid avl tree and uh we could have had you know

a tree that was just totally linear all the way down to the right it could have worked out that

way if we weren’t using avl trees uh and we would have had a really slow binary search tree but with

avl trees this is like you know a lot a lot faster reasonably and you can imagine if i added some

And you can imagine if I added some more data, let’s see, how long is this video?

Oh, it’s this video.

It’s already 15 minutes.

Let’s add one more piece of data just for fun, Zars.

I’m going to add 55.

Anybody remember that old song, I Can’t Drive, 55?

So we’re going to add the 55.

I’m going to say, here it is.

And then I’m going to do just like a new node duplicate.

node duplicate. I’m going to stick a 55 in there. And then where does the 55 belong? Well,

it’s not going to go with the root node belongs on the right side. Not going to go there belongs

on the right side, not going to go there belongs on the right side, we just end up going all the

way to the right. Pretty much. Okay, so I’m going to say it belongs there. And I’m going to connect

it. And then we update the balance factors. So zero here, and then the 42 becomes a one and the

42 becomes a 1 and the 38 becomes a 1 and the 21 becomes a 1 on the left and 3 on the right

which means we have a balance factor of 2 so now the 21 node is imbalanced this is no longer a

valid abl tree we got to rotate to make it better so once again we look at the first node that’s out

of whack or the lowest node that’s out of whack it’s going to be the 21 we call it our z and then

we find a child of z with the tallest subtree so it you know is it going to be the 12 or the 38

the 12 or the 38 it’s definitely going to be the 38 because 38 is the taller subtree same thing to

find x it’s got to be a child of y should we go left or should we go right we have to go right

because that’s the taller subtree so now we have our x y and z um i think i’ve been forgetting to

is 38, Z is 21, X, Y, and Z, and then we do A and B and C. So A is the least value, 21, B is the one

that belongs in the middle, C is the one that belongs on the right side. Now we’re ready to

draw our target output pattern. So I’m going to do this, a little duplication.

Do you like when I say, oh, we’re going to do this, we’re going to do this, we’re going to do this?

Because sometimes I just go, I don’t like it.

And I don’t think people like it when I say that.

Okay.

I’m going to connect these real fast and then change the numbers to match.

I think I made that one.

I think it’s okay.

I can never tell if it’s perfectly symmetrical.

I can’t do it.

It’s perfectly symmetric.

All right.

So A is going to be on the left.

Alright, so A is going to be on the left, that’s 21.

B is in the middle, that’s 38.

C is on the right, that’s 42.

Then we have to look for children of the output pattern that are unaccounted for.

So we’re going to do, look at the Z node.

The Z node had 12, which is unaccounted for as a left child.

So I’m going to copy the 12 somewhere.

We’ll deal with it in a second.

Then we look for the right child of 21, that’s a 38.

that’s a 38 that’s already in the output pattern so we don’t have to worry about that anymore

and then we look at the y node so y had left child was 30 that’s not accounted for so i’m

going to copy that put it somewhere else and then its right child was 42 that’s already

handled in the output pattern so we’re done there i’m going to duplicate this and then we look at

the uh we look at the x node the 42 it had a right child of 55 that also was not accounted for so i’m

for so I’m gonna just copy it over here and we’ll do something with it okay we’re done looking for

unaccounted for children again if you are just rearranging pointers you don’t really have to

worry about this in your code but since we’re doing a diagram we have to double check that

none of these nodes at the bottom here the unaccounted for children have children of their

own if they did the children would have to come with them so 55 uh no children 12 no children 30

So those are fine. If there were any children underneath, they would just stay on those nodes.

So now we have to figure out who goes where. So the 12 is the least value. That means it’s going

to go probably in the leftmost position compared to the other ones. But if we figure out like,

where does this belong in the binary search tree? It belongs right here as the left child of 21.

The 30 node, same thing. Where does it belong? It belongs between the 21 and 38. That’s the only

And then the 55, it belongs on the right side of 42 for the same reason.

Okay, I’m going to do my connecting lines real fast.

If I don’t do my connecting lines, then I don’t like it.

So now we have how many nodes in the output pattern?

We have one, two, three, four, five, six.

there’s also just six which means the final tree after rotation is just everything that we’ve

written on the right side so i’m just going to erase all of this stuff get rid of that too

and then move all these nodes over to the left call that the tree this is now the tree and then

i just have to recompute the balance factors to make sure that everything is okay so the leaves

0, 0, the 42 gets a 1, the 21 is perfectly balanced so that gets a 0.

The 38 is also perfectly balanced.

Again, it’s not about weight or mass even though there are more nodes on the left subtree,

it’s all about the height.

The height of the left subtree is 2, the height of the right subtree is 2, so 38 is actually

perfectly balanced.

Because there are no nodes that have a balance factor of 2 or worse, we are done rotating

this tree.

This is a valid AVL tree and it’s pretty fast.

So I think that’s pretty much it.

That’s all I wanted to go through today for this video.

Thank you so much for following along.

I hope you learned a little bit of stuff and had a little bit of fun.

And I will see you in the next video.

Have a great night and or day and or week and or weekend.

Hey everybody.

Thanks for watching this video again from the bottom of my heart.

I really appreciate it.

of my heart I really appreciate it I do hope you did learn something and have

some fun if you could do me a please a small little favor could you please

subscribe and follow this channel or these videos or whatever it is you do on

the current social media website that you’re looking at right now it would

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just I’ll be able to keep making videos in general so please do do me a

do me a kindness and uh and subscribe you know sometimes i’m sleeping in the middle of the night

and i just wake up because i know somebody subscribed or followed it just wakes me up and

i get filled with joy that’s exactly what happens every single time so you could do it as a nice

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and it’ll take you to my main website where you can just kind of like see all the videos i published

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if you have a suggestion for uh uh clarifications or errata or just future videos that you want to

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wake up for those in the middle of the night i get i wake up in a cold sweat and i’m like

it would really it really mean the world to me i would really appreciate it so again thank you so

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Thank you.

It should all work out according to plan, trust me.

Anyway, um, so…

The post AVL Trees: Adding Linear Data and Performing Rotations appeared first on NeuralLantern.com.

In this hands-on AVL tree tutorial, we take a massive unbalanced linear binary search tree and perform multiple rotations to turn it into a properly balanced AVL tree. Watch as we identify imbalance, label X Y Z nodes, determine A B C, handle outstanding children, and reattach subtrees step by step.

Perfect for computer science students learning data structures, self-balancing trees, and AVL rotations. We go through several rotations on the same tree to show the full process from start to finish.

If you’ve seen the basics, this is the practice video you’ve been looking for. Timestamps and clear diagrams included.

Like and subscribe for more data structures content!

00:00 Introduction to AVL Rotations Practice
00:22 Previous Videos Overview
00:56 Understanding the Linear Tree Problem
01:24 Why Balance This Tree
02:20 Computing Balance Factors
03:16 First Rotation Setup XYZ
04:04 In-Order ABC Pattern
05:50 Reattaching Subtree
07:17 Recompute Balance Factors
08:06 Second Rotation Setup
09:20 XYZ and ABC for Second Rotation
10:08 Drawing Output Pattern
11:40 Placing Outstanding Children
13:32 Third Rotation Setup
14:28 XYZ for Third Rotation
15:02 Output Pattern and Children
18:28 Recompute Balance Factors
19:28 Fourth Rotation Setup
20:04 XYZ for Final Rotation
20:32 Handling All Outstanding Children
23:20 Reattaching Final Subtree
24:50 Last Rotation Setup
25:48 XYZ and ABC Final
26:38 Output Pattern and Children Placement
30:16 Final Balance Factors Check
30:52 Valid AVL Tree Achieved
31:07 Conclusion and Thanks

=-=-=-=-=-=-=-=-=

Thanks for watching!

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Hello there! Let’s practice AVL rotations with a big ugly gross linear tree that for some reason hasn’t been rotating up to this point.

But we’re going to rotate it all at once to make it nice and balanced per the rules of AVL trees, just for practice.

Okay, so first off, you should have hopefully seen my previous videos.

search trees how to define them how to build them search through them add remove stuff all that

stuff and then we talked about abl trees which are really just self-balancing binary search

trees with some extra special rules on top we talked about the types of rotations how to do

the rotations when to rotate when to how to detect whether a tree is actually a valid abl tree and

all that stuff so if you don’t know what i’m talking about see my previous videos otherwise

we’re just going to practice on this one tree right now we’re just going to do a practice run

run. Okay so you would never I mean you would hopefully never see an AVL tree

that looks like this. This is way too imbalanced for an AVL tree. An AVL tree

would have started rebalancing itself a long time ago but let’s just pretend for

the sake of argument that you disabled the balancing feature of your AVL tree.

You maybe like have a boolean inside of your class that you’ve written in your

program called am I behaving like an AVL tree true or false and you set it to

for a while then you started throwing nodes at the tree after you ended up with this giant linear

tree then we’re going to suddenly turn on the avlness bool and start rotating this tree sucks

this is a valid binary search tree but like if you notice it follows all the rules of a binary

search tree and the data is in order but this is definitely not a valid avl tree the time

complexity of searching through this tree would be o of h and since o of h is actually the number

h is actually the number of nodes in the entire tree the time complexity of

searching through this particular tree is o of n so this is a linear tree it’s

no faster to search through than a linked list at least in terms of

scalability so this is bad we need to fix this so now let’s turn on the AVL

ness we’ll say okay now you’re no longer pretending to be a regular binary search

tree you’re an AVL tree let’s do some rotations the first thing we should do

do if you’re presented with a tree like this all at once is just compute the balance factors for

every single node. So I’m going to say the 55 is a leaf it has a balance factor of 0, the 42 has a

balance factor of 1, and really the balance factors just increase by 1 for every level up we go. So

the balance factor of the 15 node is horrible, it’s really really really bad. Now that we’ve done

this we could rotate anywhere we wanted to. I mean it’s valid to rotate starting with the 34 or the

or the 27 or 22 or 15 you wouldn’t want to rotate the 42 because that’s not

imbalanced enough for an AVL tree but really the smartest thing to do is

rotate as low as possible because when you rotate stuff that’s lower it tends

to fix stuff that’s higher so then you’ll end up doing less work and

probably end up with a tree that matches what someone else expected okay so we’re

going to rotate as low as possible that means I’m going to find the lowest node

and I’m going to say it’s the 34 node and I’ll label that as Z.

That’s bad. Let me try again.

34 is Z.

Okay, then we find the A child of the Z node

and we have to find the child that has a taller subtree,

but there’s only one choice.

We can only go to the right.

There’s nothing on the left.

So that means Y is the 42 node.

Same thing for the X node.

We’re labeling for X and we have to find a child of Y.

We have to look at the taller subtree,

We have to look at the taller subtree, but there is no other choice.

We can only go down and to the right.

So now we know our XYZ.

I’m just going to place these labels up here at the top.

I’m going to say X is 55, Y is 42, and Z is 34.

And maybe I’m going to change that text color real fast.

Then we want to produce an in-order representation of XYZ.

This step, I think it helps people with diagrams, but it also helps you a lot when it comes time to code.

when it comes time to code,

because coding for these rotations is way easier

than diagramming these rotations.

You pretty much just have to come up with three new pointers,

A and B and C,

and all they are is the ordered version of XYZ.

So, you know, which node belongs on the very left,

that’s gonna be 34,

which node would belong in the middle,

that’s gonna be 42,

which node would belong on the right side,

that’s gonna be 55.

And so now we’re ready to come up with our output pattern.

So I’m just gonna take one of these nodes

So I’m just going to take one of these nodes and copy it.

Maybe get rid of that little stem there.

And then I’m going to just, you know, copy paste this a couple of times as quickly as I can.

It doesn’t need to be perfect.

But if it’s not, they’ll laugh at you.

Okay.

So we’ll do this.

And we’ll draw some connecting lines.

So we’ve got like three nodes here.

A perfect little tri-node subtree because we picked X and Y and Z.

because we picked X and Y and Z and that became A and B and C.

Can’t get that line right, I’m going to give up.

So then we, I’m going to fix the labels here. So what is A belonging on the left?

That’s 34. What is B which belongs in the middle? That’s 42.

What is C which belongs on the right? That’s 55. So now we have

a perfectly balanced trinode subtree. Notice how this is a valid binary search tree.

If the numbers were in a different order, this wouldn’t be

a valid binary search tree and you’d have to try again with your ordering and also this has to be

a perfectly balanced trinode subtree otherwise this probably wouldn’t actually help if we if

we did a different output pattern it wouldn’t help the tree so then the next thing we need to do is

just double check that we don’t have any nodes that are unaccounted for from the input pattern

the xyz nodes so if we look at the uh the 34 node let’s look at z first i guess we’ll look at the 34

had 42 as a child but 42 is already handled so that’s actually okay then we look at the y node

that’s 42 it had 34 and 55 as children the 34 was handled uh here and the 55 is handled here so we

actually are fine we don’t need to worry about that now we’ll look at the x node the x node had

no children so we’re pretty much done again like i said in previous videos there could have been up

You know, each of these XYZ nodes could have had its own children.

Notice how 34 could have had one, 42 could have had one, 55 could have had two.

So there would be four outstanding children.

So we’re actually done at this point with the outstanding children.

There are none.

So I’m ready to reattach this tree, I guess the rotated subtree, into the regular tree.

So these three nodes can just go away from our diagram.

In code, you’re not actually removing these nodes.

You’re just sort of, you know, reattaching pointers.

reattaching pointers but I’m going to redraw this in a way that looks kind of nice and I’ll just

maybe redo this line real fast so I’m going to do a blue line here okay now we have to recompute the

balance factors so I’m going to say that the 34 is a leaf it’s got balance factor of zero same for

perfectly balanced and then we have to work our way up until we reach the root node. So

the 27 node, it’s got two nodes hanging off the right side and nothing on the left.

So its new balance factor is two. So it actually improved a little bit.

The 22 node, one, two, three, it’s got three nodes hanging off the right side and nothing

on the left. So it improved a little bit to three. And I’m guessing that the 15 also improved to four.

You can see the height on the left subtree of 15 is one, two, three, four, nothing on the left,

So we’re done with this particular rotation.

But because we see a node that has a balance factor of two or worse,

we’re not actually done rotating.

This is still an invalid AVL tree.

And under the hood, we would continually just, you know, rotate up and up and up the tree

until everything looked like it was fixed.

So again, the first node that’s out of whack, or I guess the lowest node that’s out of whack,

we’ll call that our Z node.

So we’re going to put a Z here.

piece of text and then we have to find a child of Z with the tallest subtree so there’s no child on

the left which means we have to go to the right to find our Y node okay then we have to go find X

through a child of Y and we have to find the child with the taller subtree actually the left child

and the right child have the same height of their trees of their subtrees so it doesn’t actually

know for fun let’s just go to the left because it’ll probably be harder we’ll say that the x

node is 34. be consistent if you always whenever you see like sub trees that are equal height

i don’t know maybe try to be consistent going left or right but in this case it doesn’t really matter

so we have xyz i’m going to type them up again right here we’re going to say x equals

Let me just double check that 34 42 27 and then we’ll produce ABC, which are the in order

representations of XYZ.

Again, imagine you’re using pointers if you’re doing this in code.

So we have like a is equal to the least value, the one that would belong on the left.

That’s the 27 B should be 34 and C should be 42.

And then, you know, just visually check.

way so that all the nodes have their own space. The one that’s furthest on the left is Z and that

ends up being A. The one in the middle is X, that’s B, and the one on the right is C, that’s 42, Y.

Right, yeah. Okay, so now that we’ve selected our ABC we can start drawing the output pattern. So

I’m going to like make a little copy of this and I’m just going to very quickly try to draw

balanced maybe a little bit lower they’ll laugh at you um they won’t they will i’m just kidding

let me finish this line

all right fix the numbers real fast uh so a is supposed to be on the left so that’s 27 that’s

already good uh b is 34 c is 42 so now we have to just worry about the unaccounted for children of

unaccounted for children of ABC or XYZ. So first I’m going to look at X. Whoops.

We look at X, it’s 34. It had no children of its own, so we don’t even have to worry about any of

its children. So we’re done. We look at Y, Y had two children. It had 34, but 34 was already handled

up here. And it had a right child of 55. Oh, so we have a child that is unaccounted for now. So I’m

duplicate it over here. We got to put it somewhere. I don’t know where just yet.

You can probably figure it out, but I’m just going to put it over there. Now we’re done looking for the Y.

And then next we’ll look at the Z node.

So I’m going to do

the Z node is 27. It had a 42 as a right child, but 42 was already handled in the output pattern.

So we’re done looking for outstanding children.

So the next thing I’m going to do is figure out where the heck does that 55 go?

does that 55 go? Remember, we’re trying to make valid binary search trees, which means there’s

only one place where this 55 node could actually go. If you stuck it on the left side, a greater

value on the left of 27, that’s invalid. That wouldn’t work. 55 is greater than 34, so it can’t

go on the left. It’s greater than 42. So actually 55 belongs all the way on the right side, because

again, that’s the only place it would actually work to allow us to still have a valid binary

search tree when we were done placing it so double check the subtree that you have written down after

you’re done doing all this stuff and make sure that the whole entire tree is a valid binary search

tree if not you probably have to try again okay so i’m going to duplicate this slide right here

and we’re going to say that we’ve got four nodes ready to be inserted in place of these four nodes

over here one two three four so i’m going to just select all four of these nodes and just delete them

and reinsert the rotated, you know, the rotated nodes, which are the same nodes, but they’re just,

you know, they’re drawn differently with, you know, different connections. And then I’m going

to fix this connection over here. So that’s 34 is now the right child of 22. Then I have to do

the balance factors for all of the nodes that we rearranged. So the leaves are easy. As usual,

This 42 has a balance factor of 1, and this 34 has a balance factor of also 1.

It’s not perfect, but it’s okay.

Then we work our way up the tree until we find the root node.

So now we’re going to recompute the balance factor for the 22.

Actually, it doesn’t improve.

If you notice, the right subtree has a height of 1, 2, 3.

There is no left subtree, so the balance factor is 3.

Same thing for this 15 node.

So we made a little progress.

Not much.

We have to keep going.

So, you know, once again, I’m going to select the lowest node that’s out of whack.

Or I guess as I work my way up the subtree, I can see that three node is still out of whack.

So I’m going to call that Z.

Nope.

Somebody remind me to do what types of rotations to talk about the rotation types.

Somebody send me an email or a comment.

I’m going to do it at the end of this video, but send me a comment anyway.

So the Z node is the 22 node.

22 node to find the y node we have to take a child of z with the tallest subtree but that’s obviously going to be 34

There’s no other choice

And now we actually do have an interesting choice to find the x node

It has to be a child of y, but it’s got to be the child with the tallest subtree

So for the first time we can kind of see

That we you know, we might have wanted to put x on the left or right, but it has to be on the right

We’re not allowed to put it on on the left at least for this method

So there we go. We have our XYZ. I’m going to go ahead and do X is equal to 42, and Y is equal to 34, and Z is equal to 22, and then A, B, and C are going to be the in-order representations.

the middle 42 goes on the right so now i’m ready to draw my output pattern so i’m going to like

select this real fast duplicate it up here make some copies making copies anyone no okay just

just wondering okay i’m going to do this do some connecting lines

So A is supposed to be on the left, that’s 22, and then B is going to be 34, and C is

going to be on the right, that’s going to be 42.

So we have a perfectly balanced trinode subtree.

Now we’ve got to double check for outstanding children.

So we start by looking at the Z node.

Oh my god.

Not as skilled as I think I am.

We look at the Z node.

which is already handled over here so we’re fine that way then we’ll look at the y node the y node

was 34 it had a left child of 27 that is not accounted for in the output pattern so i’m going

to grab it and then also you know when you’re grabbing these unaccounted for children double

check that they might also have children of their own if they do we have to take them with the node

that we’re moving so if 27 had any children we would not be you know removing or rearranging

children and just kind of you know let them stay in their position underneath 27. You know we would

not rearrange pointers that are further down but in this case what have I done in this case 27

doesn’t have any children I’ll have to do a copy that’s why make a copy put it somewhere so now

we’re done with the 27 on the right side of 34 it had a child of 42 but that’s already handled

42 and then I’ll just put like an arrow here. Now we’re ready to look at the X nodes children.

So the X is 42. It had a right child of 55 which is also not accounted for in the output pattern.

So I’m going to just move that over here.

Okay now we have to place these children in their appropriate positions.

So the 55 if you think about it there’s only one position that the 55 could ever go

go to produce a valid binary search tree so it can’t go there it’s got to go all the way on the

right side so that’s where we’ll put it same thing for the 27 there’s only one place it could go it

can’t go here because it’s less than 34 so it can’t be on the right side of 34 it’s going to

be on the left side of 34 and if you look at where is it going to go with respect to 22 it can’t go

on the left it has to go on the right because again we need a valid binary search tree

So I’m just going to connect this line right here.

And now we’re done making our little output tree, subtree.

So how many nodes did we have in the input tree?

It was 22, 34, 42, and 55, so like five nodes.

And if you look at the output pattern here, there’s also five nodes.

So we’re pretty much ready to just kind of remove all of these nodes

all the way up to that 15.

and make it the right child of 15. So I’m going to select all of this stuff, make it the right child

of 15, and then I guess I’ll just redo that line real fast. So this is a nicely formatted diagram.

So we’ll do what’s going on with my computer. Okay, sometimes it lags. I’m gonna, well,

balance factors now so I’m going to do balance factors for the leaves pretty easy the 22 and the

42 have a balance factor of one again if this is confusing see my previous videos we practiced a

lot of that the balance factor for the 34 is going to be a zero remember it’s about the height of the

left subtree versus the right subtree which there’s no difference actually they both have a height of

two it’s you know just just because there’s a little diagonal bounce there on one side doesn’t

awesome. Not perfect, but it’s pretty good. We look at the 15. What’s the new balance factor of

the 15? It’s actually three now because the height of its right subtree is three and the height of

its left subtree is zero. So its balance factor improved to three. So we are making, you know,

a little bit of progress. Now we know that the only node that’s out of whack in the whole tree

is the 15. So we’re just going to call that our Z node for another rotation. So I’m going to put Z

of z which would be y and there’s no other choice and then once again we have to find a child of y

to find x and we have to take the child that has the taller subtree but both of these subtrees

are equal so i’m just going to go to the right for fun i don’t know i’m going to do x over here

on the on the right side well hang on have i been doing this the easy way oh yeah let’s let’s go to

Okay, so now we have XYZ. I don’t know why I say these things. We’ll do XYZ. X equals 22. Y is equal to 34. I say them for fun. Z is equal to 15. And then we’re going to do A and B and C. So A is going to be the least value. So that’s 15. B is going to be 22. C is going to be 34.

tree so i’m going to copy this over here a couple more times this tree is probably going to be huge

i’m probably gonna have to rearrange this a few times just do this over here make another copy

over here there’s a lot of nodes that are going to be unaccounted for here um i think this was

the example where we ended up just replacing the entire tree let’s see all right i guess that’s

And then I’m going to update the numbers. So A is 15, and B is 22, and C is 34.

So now one by one we have to look for any outstanding children of the nodes in the output pattern.

So we look at, let’s say, the Z first.

The Z had a right child of 34, but that’s already taken care of in the output pattern, so we don’t have to do anything.

now we look at the y the y had a left child of 22 but that’s already taken care of in the output

pattern so we don’t have to worry about that it also had a right child of 42

um so we don’t have to worry about that sorry we do have to worry about that so i’m going to copy

the 42 over here and we’re going to have to do something with it eventually now we’re ready to

If we’re looking at the X node, X node is 22.

It had a right child of 27 that is not in the output pattern.

So that’s something that’s unaccounted for.

I’m going to place it over here.

Okay, now we’re ready to attach the nodes in the output pattern.

Let me just bring something to your attention real fast.

Notice how the 42 node, which was an outstanding child of one of the output pattern nodes,

it had a child of its own, right?

on the bottom here, you know, below the output pattern, if they had any children of their

own, we would not like change their topology. We would not assign them to different parents.

They would just stay exactly where they are, which means just for the purposes of this diagram,

I should probably also bring the 55 down as the right child of the 42. So I can remember,

you know, that it was there and not totally forget it and ruin the tree.

27 had no children, so it’s fine. The 27 should be a left child of 34 because it’s greater than

of 34 because it’s greater than 22 and it’s less than 34 so I’m going to stick it right there and

then where would the 42 go well it would it would be a right child of 34 because it’s greater than

34 so maybe I should make the 27 like just a little bit more to the left okay then I’m going

to reconnect so the 34’s left child is 27 its right child is 42 and then you know this is easy

like I make mistakes all the time too so I just want to double check that this is a valid binary

search tree because I could have you know done the numbers incorrectly so we have 15 22 27 34 42 55

it all increases it is a binary tree so it’s also a binary search tree and then it’s like

it satisfies all the other properties so now how many nodes did I actually remove four five six and

there were only six nodes in the entire tree before we did this rotation you know z is kind of the

know z is kind of the culprit of why this happened so that means the rotated pattern is actually the

final binary search tree so i’m just going to erase the whole entire tree and put the rotated

pattern in the middle here then we just have to recompute all the balance factors to make sure

that we’re done or that we’re not done so the leaves get zero and then the 42 gets a one and

then the 34 gets a one because it’s got a height of one on its left subtree and a height of two on

and a height of 2 on its right subtree, take the absolute value.

The 22 node, I don’t think that’s quite finished yet.

It’s got a balance factor of 2 because it’s here.

Let’s just clarify to make sure everybody’s on the same page.

Anytime I feel that I had to think even a little bit,

I’ll sometimes just kind of take a little step back and make sure everybody’s okay.

So the left subtree has a height of 1.

You can see that I highlighted the left subtree.

The right subtree has a height of 1, 2, 3.

So 1 minus 3, take the absolute value.

That’s going to be 2.

value that’s going to be two so the balance factor of the 22 node is two meaning we’re not done

rotating our tree it could be a little bit better a little bit faster so maybe i’m going to move this

whole thing over to the left a little bit and then we’ll call the 22 node the z node because it’s the

first node that’s out of whack then we have some more interesting choices right remember before

if we’re going to go down into the left or down into the right it always seemed obvious but now

seemed obvious but now to find our y we could go left or right but we have to go to the right

because 34 has the taller subtree 15 it’s it’s a tree of just one it’s a height of one and 34 it’s

a tree of height three so we go to the right to find y to find x again we need to look at a child

of y with the tallest subtree so that’s not going to be 27 that’s a height of one it’s going to be

42 with the height of two so the x node is this over here and now we’re ready to talk about

ready to talk about what is XYZ and ABC. So X is, I always forget to change the color before I type

this. Okay. X is 42. Y is 34. Z is 22. And then we get A and B and C. Okay, so A is the least value,

the one that would belong on the left. I mean, you can just look visually on the diagram again.

on the left right so that would have to be a because anything that starts leftmost should end

up left leftmost in terms of a well-drawn diagram so b is going to be in the middle that’s the 34

and then c is going to be the greatest value which is 42 again c 42 is the rightmost value both

before and after we decided what is a xyz and abc so we’ve got that i’m going to go ahead and start

for another new diagram.

We’re going to say,

we got that.

Give it a little copy.

Give it another little copy.

Give it a copy.

Then I’m going to draw my connections.

Dang, how long is this video?

Man, I knew this was going to be a gnarly example.

That’s why I like doing it.

Okay.

Fix the numbers real fast.

So the 27 is A, it’s 22.

And then this is going to be B, which is 34.

And then 27, or the rightmost node, is going to be the 42.

Then we start looking at children that might be unaccounted for in our target pattern.

So I’m going to duplicate this.

And we’ll say, start by looking at Z.

It had a left child of 15, which is unaccounted for in the output pattern.

So I’m going to move 15 over here somewhere.

And then we look at the 34.

That’s already accounted for in the output pattern.

So we’re done looking at Z’s children.

So I’m just going to do this.

Now we’re ready to look at Y.

Y had a left child of 27.

That’s actually unaccounted for.

So I’m going to move the 20.

I’m going to copy the 27 over here.

And then we look at the 42.

The 42 is already handled in the output pattern.

So we don’t have to worry about it.

worry about it okay now we’re ready to look at the x nodes children so we look at x it had a right

child of 55 that’s also that’s also not accounted for in the output pattern so i’m going to make a

copy so we have like three nodes to worry about and then you know just for diagram purposes in

the code this would be easy to do but for the diagram we have to make sure are we dragging any

of these unaccounted for children away from their other children that they might have so 55 it has

children 27 it has no children 15 also has no children so these are pretty easy so just visually

I’m going to rearrange I lost my camera what happened test test test whoa I’m tripping so I

had a technical issue with my camera I had to edit out I guess a little portion where I was super

confused and screaming and now I think we’re okay so we can continue the video so we’ve just looked

children that are unaccounted for and accounted for them that’s so nice of us

and we’re going to have to place them somewhere so this 15 again it can’t go

anywhere over here on the right side that would be an invalid binary search

tree it’s actually got to go on the left of the 22

it’s the only place it could possibly go same thing for the 55 where is that

going to go I don’t know maybe like all the way to the right because that’s the

only numerically valid you know place that it could go to have a valid binary

27 can’t go on the right of 34 it’s too small it ends up going on the right of 22.

So now we’ve placed the children to account for them and I’m going to just like draw my

connecting lines real fast and then how many nodes do we have here let’s see the 22

1 2 3 4 5 6 1 2 3 4 5 6 oh actually that’s every single node in the tree just like last time we

again guess that that would have happened because it might be likely to happen because the z node

here was actually the root node of the tree. So I’m just going to you know replace the entire tree

and say that we’re done rotating now. So we’ll just do this put this in the middle and then

recompute all the balance factors just to check to see if we’re done. You can probably tell already

this is a pretty decent tree so we might not have to rotate but I’m going to go ahead and

have to rotate but I’m going to go ahead and just double check anyway. So the leaves get zeros

and the 22 gets a zero because it’s got left and right subtrees of equal height. The 42 has a one

and the 34 also has a zero because it’s left and right subtrees are equal height. This is maybe

another good opportunity for me to point out that it’s not about weight or mass on the left versus

right subtrees. It’s just about height only. So even though there are less nodes in the right

The fact remains that the heights are the same. So actually that’s a zero for the root node

This is now finally a valid AVL tree. We have rotated it enough. It is now considered

Log time again, it’s going to be fast to search and insert and remove and all that stuff. So this is pretty great

We are now officially done with this example tree. So thank you so much for watching this video

I hope you learned a little bit of fun and had a little bit of stuff

and I hope you enjoyed doing this tree with me it’s always a little stressful because it’s like

a half hour tree and I always wonder like did I make a mistake five hours ago I don’t even know

so I guess I’ll see you in the next video thanks for hanging in there with me

hey everybody thanks for watching this video again from the bottom of my heart I really

appreciate it I do hope you did learn something and have some fun if you could do me a please

Could you please subscribe and follow this channel or these videos or whatever it is you do on the current social media?

It’s a website that you’re looking at right now

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So please do do me a kindness and and subscribe

You know sometimes I’m sleeping in the middle of the night and I just wake up because I know somebody subscribed or followed

subscribed or followed. It just wakes me up and I get filled with joy. That’s exactly what happens

every single time. So you could do it as a nice favor to me or you could you could troll me if

you want to just wake me up in the middle of the night. Just subscribe and then I’ll just wake up.

I promise that’s what will happen. Also if you look at the middle of the screen right now you

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as I fade into the darkness, which is coming for us all.

Thank you.

so

Hey there! Let’s rotate an AVL tree…

Hmm…

Test test test! Whoa, I’m trippin’!

The post AVL Tree Rotations Practice: Balancing a Huge Linear Tree Step by Step appeared first on NeuralLantern.com.

In this video we walk through a real example of maintaining an AVL tree by performing a rotation. We start with an unbalanced tree, compute balance factors, identify the Z Y and X nodes, and apply a double right rotation to restore balance. Perfect follow-up if you’ve seen the basics of binary search trees and AVL trees.

Watch as we turn an invalid AVL tree into a perfectly balanced one with clear step-by-step instructions.

If you’re learning data structures and algorithms, this practical example will help you understand when and how to rotate.

00:00 Introduction to AVL Tree Rotation
00:14 Prerequisites and Previous Videos
00:36 AVL Trees Overview
00:40 Types of Rotations and Balance Factors
01:01 Examining the Example Tree
01:24 Confirming Binary Search Tree Properties
01:27 Computing Balance Factors
01:39 Identifying Imbalance at Node 65
02:01 Locating the Z Node
02:54 Finding Y and X Nodes
04:08 Assigning X Y Z Values
04:33 Creating ABC In-Order Representation
04:58 Drawing the Target Rotation Pattern
05:34 Updating Node Values in Pattern
06:04 Checking Unaccounted Children
07:00 Reattaching Nodes and Performing Rotation
07:56 Recomputing Balance Factors
08:24 Updating Root Balance Factor
08:41 Resulting Perfectly Balanced Tree
08:48 Identifying Double Right Rotation
09:06 Explaining the Rotation Process
10:16 Conclusion and Next Video Teaser
11:56 Channel Promotion and Outro
15:48 Final Hello and Recap

=-=-=-=-=-=-=-=-=

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Hey there! Let’s maintain and rotate an AVL tree which is a self-balancing binary search tree.

Okay. Alright, so hopefully by now you’ve seen my other videos. If you haven’t,

you probably won’t understand this unless you’ve seen other videos elsewhere.

But in my previous videos we talked about what is a binary search tree, how to make one, how to

binary search tree, how to make one, how to define one, how to know you’re looking at one, how to build one, search,

insert, remove all the operations and time complexities.

And then we talked about AVL trees, which are a self-balancing binary search tree.

And we talked about what are the different types of rotations you can do? Why would you rotate?

How do you compute balance factors? How do you know when it’s time to rotate?

How do you do the rotation? Stuff like that. So this is just going to be one

example tree, and we’re just going to try to burn through it. If you want to learn more about

see my previous videos. Okay, so supposing now that you have a tree that looks like this, let’s

confirm that this is a valid AVL tree. This seems to be a tree. It’s a connected graph. It’s got no

cycles. It’s a rooted tree. It’s a binary tree. There are no more than two children per every

single node. And the data is in order. So 14, 22, 36, 41, 48, 55, 65. So this is a valid binary search

Now to determine if it’s a valid AVL tree, we just have to compute balance factors.

So in my previous videos, we talked about doing this.

So I’m just going to try to burn through this quickly.

I want to put zeros on all the leaves because that’s pretty easy.

This 48 gets a 1.

The 65 actually gets a 2.

So at this point already, we know there’s something wrong with our tree.

We need to rotate it.

This is not a valid AVL tree because we see a balance factor that is worse than, you know,

plus or minus 1.

61 is going to have a balance factor of, let’s see, just a 1.

And so really the only thing that we need to rotate is going to be this 65 node.

This is the node that’s out of whack.

Sometimes you see, you know, the bad balance factor in other areas of the tree,

maybe like a little bit higher and also a little bit lower.

You want to rotate as low as possible first because sometimes when you rotate low,

it fixes stuff that’s higher.

This is actually only true if the thing that you’re rotating is a direct descendant,

Not necessarily a direct child, but just a descendant of the thing that you see that is higher if you have two different nodes that are

unrelated in terms of

You know, they’re not direct descendants or ancestors of one another. They’re sort of like siblings or cousins

Then it doesn’t really matter so much which one you rotate first because they’re not really going to affect each other

but

In this case if we rotate, you know starting with the 65 there’s a chance that the 41 will get fixed up a little bit

I haven’t tried this yet, so I don’t know for sure, but I have a feeling it might.

Let’s see what happens.

So we know that the first node that is out of balance we’ll call the Z node.

So I’m just going to put a Z here.

Let me just duplicate this and we’ll say Z.

Hang on a second.

I got some dashes in my pen.

There we go.

We’ll make this the Z node.

And then we have to find X and Y.

We’ll do that backwards.

We’ll find the Y first and then the X.

them by going down one level at a time so the Y node will be a child of Z and

we’ll have to find the child that has the taller subtree in this case there’s

no choice because there’s no subtree on the right side of 65 so we can only go

left but just keep in mind if there was a subtree let’s say like that then we

would still go to the left and select that 48 as the Y node let me just write

down Y node here because it would have a taller subtree notice the left subtree

in this little example has a height of 2 and the right subtree has a height of 1

That subtree has a height of one.

So let’s just get rid of that little extra thing.

Same thing for the X.

We want to find a child of Y and we’ll always take the taller subtree.

So you know, if there was a subtree over here and maybe like another like node down there,

then we would still go to the right to get the 55 because it would have the taller subtree.

In this case, there’s only one choice.

So the X node is just going to be this 55 and that’s it.

So let me just put an X right here.

are x and y and z i’m just going to write that down real fast x equals 55 i’m going to probably

change the color here y is equal to 48 uh z is equal to 65 um and then we want to make an in

order representation of xyz and just call it abc so i’m going to say a is equal to something

and so is c so a is going to be the least value the value that belongs on the left because again

these are supposed to have ordered data like binary search trees so i’m going to put 48 here

and 55 there and then 65 here now abc is the in order representation of xyz now that we’ve done

that i’m going to draw our target pattern here so let’s see i’m going to do i’m going to do a

to connect with some blue lines because that’s just like such a cool connection.

Hang on a second.

I put that a little bit too far down.

Okay.

So we’ll do that.

And then I’m going to do another connected in line.

Okay.

I had to make a little edit there.

So I’m going to just continue making this blue line.

So I’m going to do this.

And I have to update the values now.

the values now. So this 14 on the left, that should be the A node. So we’re going to see 48,

and then the new parent is going to be 55, and then the right child is going to be 65.

Always after you fill in these values, ask yourself, have I actually drawn a valid binary

search tree? If you haven’t, meaning if the data is out of order in some way, then you haven’t

drawn a valid binary search tree, and you probably need to try again. So we’ve got 48 and 55 and 60.

is are there any children that are unaccounted for in terms of the input nodes to the pattern

that we’ve just drawn down here, the output pattern.

So let me just kind of remind you that there could have been, you know, potentially some

nodes, right?

Like we could have had a right child of 65, we could have had a left child of 48, and

we could have had two children of 55.

So there’s always a potential for four nodes that are unaccounted for.

We’ll just check them one by one real fast here.

Look at the 48 first. The 48 had a right child of 55, but that’s already taken care of in the output pattern, so we don’t have to worry about that.

Now we look at the 55. We could go XYZ if we wanted to, but I like looking at the output pattern.

55 had no children, so that’s fine. And then we look at the 65. The 65 had a left child of 48, but that’s already in the output pattern.

So at this point, all children are accounted for. We don’t really need to do that much.

that were unaccounted for they would need to be attached under the 48 and 65 somewhere and each

child would only go in one place without an there’s only one place that the child could go

without invalidating the binary search tree but that’ll be i’m going to make a harder example in

the next video so now that we’ve done this we’re ready to reattach all the nodes are accounted for

that we’re about to remove from the diagram we’re not actually going to be deleting nodes or creating

creating nodes or anything like that in the code we’re really just disconnecting

pointers and reattaching them so I’m gonna just erase this from the diagram

but keeping in mind that the nodes are not really being recreated and I’ll just

put the 55 over here where the old trinode subtree was and then I’m gonna

recompute the balance factors for all of the nodes involved so you know I just

the output pattern someone put a 0 48 also 0 because their leaves the 55 it’s a 0 it’s perfectly

balanced that’s pretty sweet and now that we’re done with the nodes that we touched we have to

work our way up to the root node we don’t have to worry about anything over here on the left

because those nodes couldn’t possibly be affected by uh by the rotation we just did because only

nodes that are above the rotation will be affected so uh the last one we you know looked at was this

basically going to say all right 55 let’s look up one more parent it’s going to be the root node

the 41 so what’s the new balance factor of the 41 node it’s actually going to be better it’s going

to be a zero so with that in mind we’ve actually created a perfectly balanced binary search tree

this is definitely a log time tree super super fast and super super cool and what type of rotation

So we had XYZ. If you don’t have the rotations memorized, keep in mind my previous videos explain all of that with some fun ways to help you remember, hopefully. Maybe some unfun ways.

So I’ll just kind of like reiterate real fast here. There are two nodes that need to be moved into position. Notice how 55 is supposed to be the new parent node, which means the 65 needs to be rotated under the 55, and so does the 48.

And so does the 48. So the 48 and the 65 both are going to be rotated into position.

That means two nodes are rotated, which means it’s a double rotation, not a single rotation.

Which node was rotated first? I always rotate the Z node first, you know, the node on the very top.

And so I’ll just use the rotation direction of that Z node.

So that’s a regular clockwise rotation.

And that’s a right rotation, so this is a double right rotation.

you can do to remember is if you go from y to z you’re going up into the right so that’s a right

rotation you could also imagine that there is some negative space kind of sitting here on the right

side a little right pocket so that’s a right rotation and i’m sure there were some other ways

that i added in the other video but i can no longer remember them so i’ll just move on we’ll

say we did a double right rotation in order to get this uh this perfectly balanced binary search tree

research tree so I’m gonna cut the video here in the next video we’re going to I

think probably rotate a linear tree you can kind of see I’ve already started to

draw one here in the next video we’re gonna look at this like giant gnarly

linear tree and we’re gonna pretend that we have the ABL rotating disabled for a

while while we built this tree and then all at once we’re gonna turn it on and

do like a whole bunch of rotations and see what the tree is gonna end up

looking like it’s gonna be pretty ugly and fun anyway thank you so much for

Anyway, thank you so much for watching this video.

I hope you learned a little bit of stuff and you had a little bit of fun.

I’ll see you in a moment, hopefully.

Hey everybody, thanks for watching this video again from the bottom of my heart.

I really appreciate it.

I do hope you did learn something and have some fun.

If you could do me a please, a small little favor, could you please subscribe and follow

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It would really mean the world to me and it’ll help make more videos and grow this community.

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to keep making videos in general.

So please do me a kindness and subscribe.

You know, sometimes I’m sleeping in the middle of the night and I just wake up because I

know somebody subscribed or followed.

It just wakes me up and I get filled with joy.

That’s exactly what happens every single time.

So you could do it as a nice favor to me or you could you could troll me if you want to just wake me up in the middle

And I just subscribe and then I’ll just wake up. I promise that’s what will happen

Also, uh, if you look at the middle of the screen right now

You should see a QR code which you can scan in order to go to the website

Which I think is also named somewhere at the bottom of this video

And it’ll take you to my main website where you can just kind of like see

All the videos I published and the services and tutorials and things that I offer and all that good stuff and

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Enjoy the cool music as I fade into the darkness which is coming for us all.

Thank you.

Hello there.

Hi there.

Hi there. Let’s maintain a self-balancing AVL tree,

which is recursive. Hello there. Let’s,

The post How to Perform AVL Tree Rotations – Step By Step Example appeared first on NeuralLantern.com.

In this video we walk through a complete example of maintaining an AVL tree. Starting with a valid AVL tree, we insert a new node that breaks the balance rules, then recompute balance factors up the tree until we find the imbalance.

Watch as we identify the X, Y, and Z nodes and perform a double right rotation to restore the AVL property while keeping it a valid binary search tree.

Perfect for students learning data structures, computer science fundamentals, or anyone preparing for coding interviews.

Previous videos cover binary search trees, AVL basics, rotations, and balance factors.

00:00 Intro to AVL Tree Rotation
00:20 Prerequisites and Previous Videos
01:06 Initial AVL Tree Example
01:18 Computing Balance Factors
03:44 Adding Node 54
04:31 Recomputing Balance Factors
06:08 Detecting Imbalance at Node 78
07:20 Identifying X Y Z Nodes
08:36 Labeling X Y Z and Trinode Pattern
10:09 Output Pattern and BST Ordering
11:03 Handling Extra Children Nodes
14:50 Reattaching Rotated Subtree
15:24 Recalculating Balance Factors
17:05 Double Right Rotation Explained
19:18 Final Verification and Outro

=-=-=-=-=-=-=-=-=

Thanks for watching!

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Hello there! Let’s maintain an AVL tree by rotating some nodes after we add a new node that messes up the whole tree.

Okay, I hope that before you watch this video you’ve seen my other videos where we talk about what is a binary search tree, how to define it, terminology, how to build a tree, search through the tree, all the stuff for the tree.

the tree all the stuff for the tree the binary search tree and then my other videos where we

talked about what is an avl tree what is a rotation what are the different types of rotations why would

you want to rotate how do you do the balance factors all that stuff happens in previous videos

so uh search the video history anyway so for now i’m going to assume that you know all that because

you watch the videos and now we’re looking at an a we’re looking at an avl tree maybe first we’re

a valid AVL tree and then we will add a new node which will probably mess the whole thing up

and force us to do a rotation to get the tree back in balance.

Okay so the first thing I’m going to do is just this is an example tree, it’s got some nodes

and I’m going to compute the balance factors for every single node just to double check.

Remember the rule that we’re using is if an AVL tree has a balance factor of two

for any node, then it’s considered invalid.

The tree itself is just a binary search tree,

not an AVL tree, until we fix the imbalance.

So the 44 node, wait, hang on, that’s actually wrong.

I don’t know why I even typed that number there.

Oh, because I was saying two.

First thing I want to do is do the balance factors

for the leaves, because they’re the easiest.

All the leaves have a balance factor of zero,

because they have no difference in left subtree

versus right subtree height.

The 50 node is also kind of easy,

because it’s left subtree and right subtree

left subtree and right subtree have the same height so that’s just a zero 17 is a little bit

more tricky it’s got one node hanging off the right side so its right subtree height is one

its left subtree height is zero so absolute value of zero minus one is just going to be one

so that’s not perfect but at the same time that’s acceptable we can just sort of move on remember

only two or worse is going to make us stop we look at the 78 again let me just uh or for the first

or for the first time we look at the 78 let me just do this the hard way to make sure everybody’s

on the same page uh for the for the balance factor of the 78 node we look at the height of the left

subtree uh which is two and then we look at the height of the right subtree which is one

and so if you take the absolute value of two minus one that’s going to be one

so actually the balance factor of the 78 node is okay this is maybe a good time to point out the

We don’t care that there are more nodes on the left. We just care about the height of the left subtree

So I’m going to duplicate this slide to continue so far so good every node seems to be okay

We’ll look at the 44 node. You can probably eyeball it because

Well, I mean for me when I want to eyeball a balance factors

I just look at the left subtree versus the right subtree notice how there’s like one level of difference

in height we could do it the hard way if we wanted to and

do it the hard way if we wanted to um so like the left subtree has a height of two right subtree has

a height of three so two minus three absolute value that’s going to be one or just that one

line that i drew before which was the the difference so that means this whole tree is

actually okay it is a valid avl tree nice video over i’m just kidding uh let’s add a node that

So I’m going to add the number 54.

Let me just try to select an existing node here, duplicate it.

I’ll change this to a 54.

Then we’ll figure out where the 54 node actually belongs.

Hopefully you’ve already become an expert at binary search trees.

So I’m just going to do it real quick.

We show up top here, 54 belongs on the right.

We look at the 78, it belongs on the left.

We look at the 50, it belongs on the right.

We look at the 62, it belongs on the left.

right there as the left child of the 62 node. So I’m going to go ahead and do my little connecting

line indicating that the nodes are now pointing to each other. The 54 thinks its parent is 62.

The 62 thinks its left child is 54. And now we have to recompute the balance factors.

So obviously the node we just added, it doesn’t have a balance factor, so we have to recompute

way up the tree until we hit the root node, recomputing all balance factors along the way.

We don’t actually, I’m going to do marks here to make sure that I don’t forget.

We don’t actually have to recompute, you know, the 17 or the 32 or the 88 because those nodes

are not possibly going to, or actually even the 48, because those nodes are not possibly going

to be affected by the 54 node because balance factors are only affected by nodes that are

or even nodes below wouldn’t be affected. So we’re only going to compute the 62, the 50,

the 78, and the 44, working our way up to the root node. Soon as we hit root, then we’re done

recomputing. So the balance factor for the 62, I’m going to eyeball that. That’s going to be a one

because left subtree height is one, right subtree height is zero. So far, so good. As soon as we

away. Again as I said in a previous video we probably will see that the root

nodes balance factor is bad but if we if we compute the whole tree and then just

decide I don’t know like the root node is going to be rotated we’ll probably

rotate incorrectly you want to rotate as low as possible first because rotating

lower in a tree is probably going to fix nodes that are higher. So recomputing the

balance factor for the 50 that’s going to be a 1 because left subtree height of

right subtree height of 2, take the absolute value of the difference there.

We go up to the 78 node and this one is going to be bad because you can see that the left subtree

has a height of 3 and the right subtree has a height of 1 so the bounce factor is going to be

a 2. It’s at this point since we just started by inserting the 54 and worked our way up that I

would stop and do a rotation but just for the purposes of this tutorial I’m going to compute

just to show you what’s going on.

The height of the left subtree off the root node is two,

the height of the right subtree is one, two, three, four.

So that means the balance factor of the 44 node

also has a two.

So at this point, it might be a little confusing

if you don’t remember that you kind of have to stop

as you’re working your way up

or at least rotate the lowest possible node first.

You might be confused, you know, which one do I rotate?

Do I rotate the 78 first or do they rotate the 44?

If we were to rotate the 44 first, the root node, you know, we could do it.

But at the same time, there’s a chance that it won’t actually fix the 78 node.

And there will be other stuff underneath that is bad.

And so it’ll cost us more work to do it that way.

So instead, we’re going to stop at the 78 and do our rotation.

So remember we said in a previous video that we’ll choose the first node that is out of whack as the Z node.

We’re looking for X, Y, and Z.

I’m going to write X, Y, and Z here.

X, Y, and Z.

Now, how do we choose Y?

We go down to a child of Z,

and it has to be the child that has the taller subtree.

Notice how the left subtree of the Z node has a height of 3,

and the right subtree has a height of 1,

which means we definitely need to go left to find the Y node.

So I’m going to put Y right here.

If the left and the right subtree had the same height,

then it doesn’t really matter which one you choose.

matter which one you choose. Maybe try to be consistent, but definitely we have to go left

this time because that’s the taller subtree. Again, look how tall it is. Okay, now we need to look for

the X node. Again, we have to take the left or the right child of the Y node, or just like a grand

child of the Z node. That is definitely a child of the Y node. We have to choose the taller subtree,

so we could not put X as 48. We can’t do that because that’s not the taller subtree. Instead,

Instead, the X is going to be the 62 node because that’s the taller subtree.

Now we have chosen our X, Y, and Z node,

which means I’m going to just type some stuff up here real fast.

I’m going to do X is equal to 62, Y is equal to 50,

and Z is equal to 78, just as a reminder.

And this is kind of similar to what you would do in the code.

I mean, on these diagrams, us humans,

we have to like rearrange things and name the types of rotations and so forth in the code it’s

actually really easy uh in comparison because all you have to do is produce an in-order representation

of the three node pointers that you received x y and z let’s say we have a function called rotate

we don’t even have to remember the types of rotations you just grab xyz and then you order

them and call them a b and c let’s do a equals something b equals something c equals something

So A is going to be the least value because again, this has to be a binary search tree

before it can be an AVL tree.

So 50 is going to go on the left.

That’s the only place that it could go.

And then 62 in the middle and then 78 on the right side.

And then I’m going to just, maybe I’ll copy and paste this here.

This little trinode subtree.

Was that, will that work?

Yeah.

So I don’t have to draw this again.

I’m going to erase some of this stuff.

Oh, come on, man.

I’m going to erase some of this stuff just to make it a little more neat.

just to make it a little more neat.

Then I’m going to replace the values in these nodes with just X, Y, and Z.

So, sorry, ABC.

So A is going to be 50.

That’s the leftmost node or the least value.

And then B is going to be 62.

And then C is 78.

Before you continue, make sure you’ve actually drawn a valid binary search tree,

because if you haven’t, then your AVL tree is not going to end up being valid either.

not going to end up being valid either it always has to be a valid binary search tree so the order

has to be correct so if you were thinking about well how can we can’t put the 78 in the middle

or how come we can’t put the 62 on the left what would have happened well it would have been invalid

or you would have ended up with like a straight line or something that wasn’t the target pattern

for a rotation which is supposed to be a perfectly balanced trinode subtree okay so now we’ve got x y

here which is going to basically take away from these three nodes the next thing we need to do is

take into account any nodes that come underneath nodes from the output pattern so our xyz i guess

you could say input pattern too let’s look at uh you know x for example so like the 62. did it have

any children that are not accounted for in the output pattern actually yes it has a 54 node so

this 54 needs to go somewhere we can’t just like erase it from the tree so i’m going to move it

to move it down here just so I don’t forget it. Then we look at the Y node. What children did the

Y node have? It had a 48 as a child and a 62 as a child. The 62 is actually already handled in the

output pattern so we don’t need to worry about that but the 48 was not handled so we got to put

that somewhere too. So I’m going to move that down here. Now we look at the Z node. The Z node

had 50 as a child, as a left child. That’s already handled in the output pattern so we don’t need to

don’t need to worry about that it also had 88 and that is not in the output pattern so we have to

do something with 88 so i’m going to move that down here too i should also point out um i think

in a future video i’m going to do a more complicated example where some of these nodes that are

children of the uh the output pattern they might have children of their own or descendants of their

own if they do don’t rearrange any of the descendants so anything that was below 54 just

hanging exactly where it was already off of the 54.

Same thing for the 48.

If it had any children of its own,

just leave them in place.

Don’t touch them.

Don’t touch any descendants.

So now that we’ve done this, whoops, dang it.

Let me copy this in a good way instead of a bad way.

Okay, let me erase this real fast.

X, Y, and Z.

Nope, that’s going to screw me up.

I’m going to forget.

Okay, now we have to place these nodes

nodes of the output pattern in their appropriate position like where do they

actually belong well remember this is supposed to be a binary search tree so

there’s actually only one place that the 54 node could ever go and still give us

a valid binary search tree so I mean think about it for a second could it go

on the left of 50 no could it go on the right of 62 no it has to go in between

the 50 and the 62 otherwise we wouldn’t have a valid binary search tree so I’m

there maybe a little bit lower to make this a more beautiful diagram then I’m going to do my

connecting line real fast so I’m going to do this so it’s the right child of the 50 now notice how

before it was the left child of the 62 so then you know again we’re just really disconnecting

pointers and then rearranging them so this 48 it can’t be on the right side of 62 it has to be on

the left side it can’t be on the right side of the 50 it has to be on the left side so the 48 must be

side so the 48 must be the left child of the 50. again you must end up with a valid binary search

tree or you’ve done something wrong so i’m going to just kind of do this here

um and then same thing for the 88 node there’s only one place this could go it can’t go on the

left of 62 it has to go on the right of the 78 it couldn’t go on the left of the 78 that would be bad

news so i’m just going to place this here and then uh just make a little connecting line okay

Okay, now I’m going to double check to make sure that I didn’t forget any other nodes

because when you’re working with a diagram, you might forget a couple of nodes and just

sort of like erase too many nodes without putting them in the output pattern.

If you’re rearranging pointers, it’s fine.

You wouldn’t have to worry about any nodes that came underneath these nodes because they

would just remain attached in place.

But I just want to double check real fast.

So we got X, Y, and Z.

So that’s three nodes and then four, five, six.

So we’ve got six nodes total if we include the XYZ nodes and all nodes that are direct children of those and even underneath.

So how many do I have in the output pattern?

1, 2, 3, 4, 5, 6.

So I’ve got the same number of nodes.

I’m ready to erase all this stuff up top and reattach.

1, 2, 3, 4, 5, 6.

Yeah, reattach this into the diagram.

So I’m literally just going to erase all this stuff right here.

Then I’m going to select all the nodes in the output pattern.

the output pattern and i’m going to move them up to be the new right child of the 44 because that’s

where originally the z was so i’m just going to maybe move these up here real fast and then get

that 32 back down where it belongs dang you 32 uh actually it’s a little bit too low get that

So the last step is we have to recalculate the balance factors to make sure that we have a valid AVL tree.

Maybe we have to do another rotation, maybe not, I don’t know.

Kind of looks like maybe not, but I’m not going to take any chances.

So let’s do the leaves first.

Remember, we have to recompute the balance factors for every node that we just moved.

So all the nodes in the output pattern and their immediate children, not anything lower.

all the leaves are just going to have a balance factor of zero that’s pretty easy.

The 50 is perfectly balanced nice so it’s going to have a balance factor of zero.

The 78 has one node hanging off the right and nothing on the left so it’s got a one there.

The 62 again we’re not looking at mass we’re not looking at weight we’re looking at the height of

the left subtree versus the height of the right subtree so this is actually perfectly balanced

because the height of the left subtree is 2 and so is the height of the right subtree.

we’re done with all the nodes that we just moved we should recursively or just kind of like walk

our way up the tree and recompute all balance factors until we find the root node. Okay so

there’s the root node only one more thing left to do. The balance factor of the root node is let’s

see left subtree has a height of two right subtree has a height of three so the balance factor for

the root node is going to be a number one. Notice how the 44 the root node got fixed because we

we rotate it a little bit lower.

And this is not the fastest tree in the world,

but it is considered a log tree

because the imbalance is only scaling

with a constant factor because we’re limiting

the balance factors to just the number one basically

before we rotate.

So what type of rotation did we actually perform?

If you looked at my previous video,

you should know this already,

but maybe I’ll duplicate this here

sort of start to maybe annotate it. I’m going to do…

Okay, so what is the rotation we did? Well, if we look at the ZYX here, the XYZ,

there’s a little bit of a bounce. Actually, let me do orange. There’s a little bit of a bounce,

right? So that’s going to be a double rotation. Or you can imagine that we had to move two nodes

and the 50 on the left and the 78 on the right. Notice how the 78 had to be rotated first.

So it’s like under the 62. And then the 50 was rotated second underneath. I mean,

we snipped off that 54 elsewhere. But long story short, there are two nodes that need to be rotated

into position for output pattern. So that’s a double rotation. So diagonal bounce or two nodes

means a double rotation and then is it a left or a right rotation well if you

think about the first node that we rotated it was the 78 what’s going on my

computer the first node that we rotated was 78 the Z node so what direction did

we rotate that that was clockwise so that’s a right rotation so a double

right rotation another way of looking at it is if we only look at the input

look at the input pattern we just look at you know these nodes right here there was kind of a

pocket of empty space on the right side wasn’t there so that’s why it’s a double right rotation

another way to think of it is I’m running out of colors if we started at the y node and worked our

way up to the z node we went up into the right so that’s a double right rotation also either way you

cut it we performed a rotation and now we have a pretty good tree okay let me just double check

Okay, let me just double check.

This is a valid BST.

Doop, doop, doop, doop, doop, doop, doop, doop, doop.

And it’s a valid AVL tree

because all the balance factors are okay.

Yeah, and in future videos,

we’re going to do more rotations.

I’m probably going to do at least like,

you know, two or three more videos

where we just simply rotate, you know, some tree

and explain it and work our way through it.

If you have an idea for a really gross tree

that you want me to rotate in front of you,

leave a comment.

leave a comment if i get enough comments maybe i’ll do it um otherwise thanks for watching this

video i’ll see you in the next one i hope you learned a little bit of stuff and had a little

bit of fun i’m outie i’m just simply outie hey everybody thanks for watching this video again

from the bottom of my heart i really appreciate it i do hope you did learn something and have some

please a small little favor could you please subscribe and follow this channel or these videos

or whatever it is you do on the current social media website that you’re looking at right now

it would really mean the world to me and it’ll help make more videos and grow this community so

we’ll be able to do more videos longer videos better videos or just i’ll be able to keep making

videos in general so please do do me a kindness and uh and subscribe you know sometimes i’m

And I just wake up because I know somebody subscribed or followed.

It just wakes me up and I get filled with joy.

That’s exactly what happens every single time.

So you could do it as a nice favor to me or you could troll me if you want to just wake me up in the middle of the night.

Just subscribe and then I’ll just wake up.

I promise that’s what will happen.

Also, if you look at the middle of the screen right now, you should see a QR code,

which you can scan in order to go to the website, which I think is also named somewhere at the bottom of this video.

where you can just kind of like see all the videos I published and the services

and tutorials and things that I offer and all that good stuff and if you have

a suggestion for clarifications or errata or just future videos that you

want to see please leave a comment or if you just want to say hey what’s up

what’s going on you know just send me a comment whatever I also wake up for

those in the middle of the night I get I wake up in a cold sweat I’m like this

It would really mean the world to me.

I would really appreciate it.

So again, thank you so much for watching this video.

And enjoy the cool music as I fade into the darkness, which is coming for us all.

Thank you.

Hey there, let’s uh…

Bart on ourselves.

Hello there, let’s rotate an AVL tree because it’s frickin invalid.

No.

Hey there, let’s uh maintain an AVL tree.

and eh.

The post AVL Tree Rotations Tutorial: Fixing Imbalance After Adding a Node appeared first on NeuralLantern.com.

Hey everyone, in this video we break down the four types of rotations you need to know for AVL trees – self-balancing binary search trees.

We cover single right rotations, single left rotations, double right rotations, and double left rotations with clear diagrams and explanations of the input patterns and how they become the balanced output pattern.

If you’ve been learning about binary search trees and want to understand how AVL trees maintain their balance through rotations, this is the perfect next step. We look at the trinode subtrees, balance factors, and exactly how to rearrange the pointers for each case.

Great for computer science students, coding interviews, or anyone building their data structures knowledge.

Watch the full series on binary search trees and AVL trees for more.

00:00 Introduction to AVL Tree Rotations
00:14 Previous Videos Overview
00:28 Balance Factors and Trinode Subtrees
00:50 Identifying Imbalance
01:56 When to Rotate
02:20 Four Input Patterns
02:41 Target Balanced Pattern
04:18 Single Right Rotation
04:33 Single Left Rotation
08:44 Double Rotations
11:51 Remembering Rotation Types
13:04 Wrapping Up Rotations
13:27 Thanks and Outro

=-=-=-=-=-=-=-=-=

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Hey there, let’s talk about rotation types in an AVL tree, which is a self-balancing binary search tree.

Okay, so hopefully you’ve watched my previous videos. If you have not, you probably want to check them out first,

where we talk about what is a binary search tree, how to build one, how to define one,

terminology, searching through it, all that stuff. And then we talked about an introduction to AVL trees,

trees which are self-balancing binary search trees this video is going to be

about the types of rotations that you can perform in an AVL tree what do I

mean by rotation well the basic idea that we talked about in the last video

was that throughout your tree you’ll notice that different nodes have

different balance factors which means how imbalanced or balanced they are when

we identify a node that has a balance factor that’s too bad then we consider

AVL tree and we have to stop and rotate a couple of nodes you know kind of rearrange them to make

them better before we have a valid AVL tree so we have four different input patterns that we can

have so imagine you know this little trinode subtree up on the top left that could be you know

three nodes that we found somewhere in our tree that had a bad balance if you just quickly compute

the balance factor for this trinode subtree you’ll notice actually let me just do it real fast for fun

that the 65 node has a balance factor of 2 because its left subtree has a height of 2

and there is no right subtree. The 55 has a balance factor of 1 and 48 being a leaf, at

least in terms of this trinode subtree, has a balance factor of 0. All of these trinode

subtrees are going to have the same situation. The one on top is going to be 2 and so forth.

So you can imagine that you were going through the whole tree, computing balance factors like

And you found a two, which remember we said in our last video, two or worse, meaning two or higher,

if we’re using balance factors that are absolute or negative two and also positive two or worse,

if we’re not using absolute balance factors, means you got to stop and try to rebalance that trinode subtree.

So we found a node that had a balance factor of two and we realized we have to stop and rotate.

The pattern of the trino subtree before we perform the rotation could be one of these four things.

It could be the thing on the left on top, the thing on the right on top, the thing on the bottom left, and the thing on the bottom right.

And all four of these patterns should end up becoming, after we do our rotation, they should end up becoming the pattern in the middle.

So this is always going to be the target pattern.

Let me draw some arrows real fast.

pattern is going to end up looking like the output pattern in the middle.

Why is that?

How is that possible?

Well, we only have three nodes.

If you look at all the input patterns, they’re all the same numbers.

What we’re really doing is disconnecting all the pointers, disconnecting the parent child

relationships and then reconnecting them in the best possible way to make, you know,

the most balanced trinode subtree.

Notice how the height of all of these input trees is always three, right?

because it’s a trinode subtree that’s really really bad and out of whack so we’ll you know

like one two three one two three right height of three we’ll do a rotation rearrange the pointers

rearrange the parent-child relationships so that the final height ends up being actually just two

and if you think about reducing the height we’re making the tree better we’re making it more

balanced we’re making it faster to search through and add and remove things from so we’re making a

So we’re making a better tree by reducing the height a little bit.

If you had a really, really, really bad tree and you just went around to all the bad nodes

and just started rotating them all, you’re reducing the height by one for each rotation.

Eventually, when you get up to the top of the tree, you’ll have an overall AVL tree,

which is pretty fast.

We should say now that AVL trees are always going to be log time trees because the imbalance

factor is limited by a constant number.

We’ll say that again in a future video.

in a in a future video so anyway i’m going to label these trees now

i’m going to say that the one on the top left is a single right rotation

and the uh the tree on the top right you can probably imagine is a single left rotation

let me just duplicate this real fast so i can come on man can i duplicate this yeah so i can

edit the text real fast. This is going to be a single left rotation. And then the trees at the

bottom, those are going to be double rotations. Let me just write over here, maybe next to that

one, double right rotation. And then I’ll duplicate this. And this one over here is going to be a

these on different sides than I usually do because I was trying to make it look a little more tricky

but it’s really not that bad. So how do we know that the top pattern is a single right rotation

and not a double right rotation or it’s a single right rotation and not a single left rotation?

The first thing to try to understand is what’s going to happen between the input pattern and

the output pattern. So if you look at the input pattern here we’ve got like a straight line and

And the output pattern is like a perfectly balanced trinode subtree.

Notice how the 55 is the new parent.

And it’s also in the middle for the output pattern.

And then for the input pattern, the 55 is also in the middle, but it’s not the new parent.

But it is the parent already of the 48 node.

Notice how it’s the parent already of one node, the 48 node.

So that means its relationship to the 48 node, the left child, doesn’t actually change.

bit more you’ll realize that only one node actually needs to be moved into position so we’ll call this

a rotation try to imagine visualize maybe like your point your left pointer finger being placed

on the 55 node and your right pointer finger being placed on the 65 node and then just rotate

the 65 node downward we rotate that clockwise and the physics people out there love to say that

are right rotations.

At least as far as I remember.

Not totally sure about that,

but I’m just going to say it anyway.

So notice how the 65 needs to be rotated

to the right or clockwise.

And only one node actually needs to be rotated

to produce that final output pattern.

That’s why we call this a single right rotation

because there’s only one node to rotate.

The next thing I’ll do is just try to give you

a few extra ways of being able to remember

being able to remember whether it’s a left or a right so for starters one node needs to be rotated

so it’s a single rotation and we’re rotating clockwise so it’s a right rotation but you can

also do this you can think of it like can i get rid of that real fast you can think of it like

well there is a bunch of empty space kind of sitting here on the right side so that could be

another way to remember that this is a right rotation also if you were to start at the node

and work your way to the node at the top,

you know, the Z node,

then we’re kind of going up and to the right.

So that’s another way to remember

that this is a right rotation.

Single, you can remember that also

by just the fact that this is a straight line.

Like notice how the nodes are kind of in a straight line.

They don’t do a zigzag or a bounce.

So single rotations are rotations that look straight at first,

or I guess input patterns that look straight

and only one node actually needs to be rotated.

um right versus left is there’s a pocket of empty space on the right side and you go up and to the

right for the right rotations similarly you can probably already figure out by now that

the left rotation just means well we’re going let’s see we’re going up and to the left to get

to that top node or the only node that has to be rotated is that 48 node and it’s going to be

rotated counterclockwise which is left so again only one node needs to be rotated and we’re going

to be rotated and we’re going to rotate it counterclockwise which is a left rotation so

one node rotated single counterclockwise left or just like a pocket of empty space sitting on the

left side okay so then for the double rotations the first thing you can look at is the idea that

there’s a little bit of a bounce you know there’s like a angle bounce there if you see a bounce or

sick a couple weeks ago then that’s a double rotation another way you can try to remember it

as two nodes need to be moved into position so if I maybe I should copy this to another slide real

fast let me copy this and we’ll do another slide okay so you know which nodes need to be moved into

position let me copy also the output pattern just as a quick reminder if we wanted to make that

If we wanted to make that output pattern, let me put it up here maybe,

then obviously the 55 still has to be in the middle

because it was already in the middle before we rotated.

That means both the 48 and the 65 need to be rotated underneath the 55.

So that means we’re going to do the 65, we’re going to rotate it clockwise underneath the 55.

So I’m going to do like a little circle here, a connector, and then I’ll do like 65 there.

65 there and then after we do that rotation we have to rotate the 48

underneath the 55 so that’s going to be a rotation that is counterclockwise so

I’m going to do this circle that there and then do a 48 notice how when we’re

done doing the rotations we have a valid binary search tree you should always

check your rotated pattern your output pattern to make sure it’s actually in

binary search tree because if it’s not let’s say the numbers are out of order for some reason from

left to right then you have rotated incorrectly and you have to double check yourself so two nodes

need to be rotated into position therefore it’s a double rotation or you see a diagonal input pattern

double rotation how do you remember the difference between left and right well um if you start at the

node in the middle just like we did in the last slides we start at the node in the middle and go

and go to the node at the top, the Z node,

we’re going up and to the right.

And so this is a right rotation.

So it’s a double right rotation.

Another way to remember it is you should probably try to rotate the Z node first.

The node that’s on very top is the 65.

Rotate that first and then call that the direction of the rotation.

So, you know, we rotate the 65 first.

That’s a clockwise.

So that’s why it’s a right rotation.

Then just ignore the direction of the other node that you rotated.

you can also kind of imagine taking the original node if you don’t if you don’t

want to think of like actually rotating let me just kind of erase this real fast

you can kind of imagine that you are you’re sort of taking the new parent

that’s supposed to be the new parent and just sort of pulling it up

I draw a picture of Batman and he shoots a batarang at it and he sounds like Christian

Bale and he’s like, where is the rotation?

I’m not going to do that.

Maybe if there are enough comments, I’ll do a special video just for my really, really

bad drawing.

But imagine the 55 is getting pulled up and it’s getting pulled up past two other nodes.

So that’s why it’s a double rotation.

Another way to remember.

The 55 would be the mob boss that’s hanging by his ankle.

anyway so uh we we kind of know how we can remember the difference between uh single and

double rotations uh we’ve got a single right rotation a single left a double right a double

left there’s only going to be four possible input patterns and they’re all going to look like this

one output pattern because really what we’re doing is we’re just we’re not you know creating

any nodes or removing any nodes we’re just disconnecting pointers we’re disconnecting

just get this over here we’re saying okay input pattern I disconnect the pointers I’m saying the

65 no longer has a left child the 48 no longer has a parent the 55 no longer has a parent and

the 48 no longer has a right child and then when we’re done with that we just kind of like reconnect

the relationships we’re like okay you know what just kidding you all have parents and children

again we’re just rearranging them okay let me check out my thing real fast here

I’m going to cut the video here because this is the basic idea for the different types of rotations that you could use.

In the next videos, we’re going to actually start rotating bad trees to make them work.

So thanks for watching this video.

I hope you learned a little bit of stuff and had a little bit of fun.

I’ll see you in the next video.

Okay.

Hey, everybody.

Thanks for watching this video again from the bottom of my heart.

I really appreciate it.

it. I do hope you did learn something and have some fun. If you could do me a please, a small

little favor, could you please subscribe and follow this channel or these videos or whatever

it is you do on the current social media website that you’re looking at right now. It would really

mean the world to me and it’ll help make more videos and grow this community. So we’ll be able

to do more videos, longer videos, better videos, or just I’ll be able to keep making videos in

and subscribe. You know sometimes I’m sleeping in the middle of the night and

I just wake up because I know somebody subscribed or followed. It just wakes me

up and I get filled with joy. That’s exactly what happens every single time.

So you could do it as a nice favor to me or you could you could troll me if you

want to just wake me up in the middle of the night just subscribe and then I’ll

just wake up. I promise that’s what will happen. Also if you look at the middle of

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to the website which I think is also named somewhere at the bottom of this

is also named somewhere at the bottom of this video and it’ll take you to my main website where

you can just kind of like see all the videos i published and the services and tutorials and

things that i offer and all that good stuff and uh if you have a suggestion for uh uh clarifications

or errata or just future videos that you want to see please leave a comment or if you just want to

say hey what’s up what’s going on you know just send me a comment whatever i also wake up for

the night I get I wake up in a cold sweat I’m like it would really it really mean the world to

me I would really appreciate it so again thank you so much for watching this video and um enjoy

the cool music as as I fade into the darkness which is coming for us all

Thank you.

Hello there.

Let’s talk about rotation types in an AVL self-balancing binary search tree.

The post AVL Tree Rotation Types Explained for Self-Balancing Binary Search Trees appeared first on NeuralLantern.com.

Learn AVL trees in this beginner-friendly introduction. We cover balance factors, why regular BSTs get slow, and how AVL trees stay balanced with rotations. Great for coding interviews and data structure understanding.

00:00 AVL Trees Introduction
00:00:28 Problems with Regular BSTs
00:00:56 AVL Tree Balance Rule
00:02:10 Balance Factor Explained
00:02:48 Computing Balance Factors
00:03:11 Example Tree Analysis
00:04:40 Imbalance at 65 Node
00:05:08 Invalid AVL Tree
00:06:18 Linear Tree Problem
00:06:50 Trinode Subtree
00:07:40 Selecting Z Y X Nodes
00:09:20 Rotation Overview
00:10:03 Next Videos Preview
00:11:09 Thank You and Subscribe

=-=-=-=-=-=-=-=-=

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Hello there. Let’s talk about a type of self-balancing binary search tree called an AVL tree.

Okay, so by now I hope you’ve seen my other videos that I’ve posted

previously where we talked about binary search trees, how to define them,

terminology, how to know you’re actually looking at one, how to build them, insert,

If you finish those videos already with me, then you know that binary search trees can actually get slow sometimes because if we have bad data, regular binary search trees don’t really know how to rearrange themselves.

They just might become slow depending on how good or bad the data is.

The data is totally random.

Then on average, it’ll, you know, the trees will be log time, but that’s not always the case.

Maybe sometimes you have bad data, sorted data, poison data, whatever kind of data.

The secret to AVL trees is we’re going to start with the binary search tree.

And so first it has to satisfy all the rules of the binary search tree.

And then we add on another rule.

The rule is going to be that we don’t have a valid AVL tree.

If the balance factor for any of the nodes, we’ll talk about balance factors in a second.

If the balance factor for any node is two or worse.

Meaning if we think the tree is too imbalanced at any node, it’s not a valid AVL tree.

and therefore we must rebalance the tree with something called a rotation.

After we do enough rotations and we see that the tree is balanced again, then it’ll be valid.

And so AVL trees sometimes are invalid in some intermediate state.

Like if you have an AVL tree and you add some data into it,

for a moment it might be an invalid AVL tree and then internally it’s sort of just like rotating itself and rebalancing itself.

Then when the tree is done, you’ll have a valid AVL tree again, if that makes sense.

AVL tree again if that makes sense. So suppose this tree here we can tell this is a valid binary

search tree because it follows all the rules that we talked about before. So let’s start implementing

the first rule of an AVL tree which is let’s compute the balance factors for every single node.

How do we compute balance factors? The basic idea is I want to say BF for balance factor

the left subtree height minus the right subtree height.

If you don’t know what a subtree or height is,

you should probably check out my other videos right now.

But basically, we’re just going to take the absolute value

just to get the difference in the height of the left

and the height of the right.

So if you know binary search trees already,

then you know that the leaves, they don’t have any subtrees.

So their balance factor is going to be pretty easy to compute.

It’s just going to be a zero.

I should also point out that some other tutorials out there use positive and negative numbers

for the balance factor.

That’s okay.

I’m just going to use absolute values here because it’s a little bit more simple.

So all the leaves get a balance factor of zero, which is fine.

We’re looking to see if we can find a balance factor of two or worse to indicate to us that

we have an invalid AVL tree.

So far, so good.

Let’s look at the 22 node.

we’ll compute the height of the left subtree versus the height of the right subtree.

So, you know, I’m going to highlight its left subtree.

That’s just the 14 node that has a height of 1.

And then its right subtree is just the 36 node that also has a height of 1,

which means the balance factor for the 22 node is the absolute value of 1 minus 1 is 0.

So actually the 22 node is perfectly balanced.

If you think about it, that makes sense.

Left and right subtrees have the same height.

No problem.

have the same height no problem so I’m going to move on let’s compute the balance factor for the

48 node this is a little trickier notice how the right subtree has a height of one and there is no

left subtree which means its height is zero so for the 48 node it’s actually going to be

the absolute value of zero minus one or just one so we find our first node that’s a little bit out

imbalance here but avl trees tolerate imbalances of one they don’t really care we sort of try to

make a trade-off between constantly always rotating every single time anything happens

which would perhaps burn a little too much cpu versus letting the tree just be imbalanced to a

reasonable amount so we could still call this a log tree in the end so we consider this to be okay

65 node and compute the balance factor so 65 node has no right subtree so the height is zero there

and its left subtree is those two nodes on the left so that has a height of two which means the

balance factor for the 65 node is going to be the absolute value of two minus zero is two at this

point we are already certain that this is not a valid avl tree and it would need to be rebalanced

you know, consider it a valid AVL tree in the future.

So invalid AVL tree, still a valid binary search tree.

Something has to be done.

Normally what I would say is if you find some nodes

way down lower in the tree that are imbalanced,

then just go ahead and perform a rebalancing

or a rotation immediately,

because sometimes when you rotate lower in the tree,

you’ll end up fixing nodes that are a little higher.

But since we started with this tree

and we’re not kind of, you know,

building a tree step-by-step,

I just want to compute the balance factor for all of the nodes at the same time.

So we’ll do the same thing for the 41 node, the root node.

Its left subtree has a height of two and its right subtree has a height of three.

So its balance factor is the absolute value of two minus three is one.

So actually the root node is okay.

If it was only up to the root node, we would say this is a valid AVL tree

and we don’t really need to do anything.

need to do anything however we’ve already seen that the 65 node is invalid so again uh we we

need to do something here’s another quick example before we move on to actually identifying which

nodes to to modify and rotate just real fast i want to show you you know that a binary search

tree could actually end up being a linear tree if you had really really really bad data the binary

This is slow because this is a linear tree.

The time complexity of searching through this tree is linear time.

It scales linearly with the number of nodes in your data set.

That’s really far away from log time, which is supposed to be lightning fast.

So an AVL tree will fix this kind of bad data.

Let’s take a step back here and focus your attention for a second on these three nodes.

We noticed that there is one node that’s actually out of whack.

It’s the 65 node, right?

What we’re going to do is we’re going to find a trinode subtree that starts with the first node or the lowest node we can find that’s out of whack.

It’s the 65 node.

And you can tell that if we just kind of go down a level and down another level to select the other two nodes,

it’s just going to be this little, you know, subtree of three nodes or otherwise known as a trinode subtree.

So here’s kind of a diagram for that.

Whoops, forgot to delete that earlier.

Let me get rid of that.

So this is the trinode subtree that we selected, right?

So we’ve got the 65 and the 48 and the 55.

How did we actually select this?

So what you kind of want to do is the first node that is out of whack,

you want to call that Z.

So I’m going to write a Z here.

What we’re looking for is a trinode subtree,

which we’ll first call X and Y and Z.

And then we’ll end up reordering the nodes

and then rearranging all the pointers to the nodes

so that the trinode subtree is a little bit more in balance.

more in balance. So we see the Z node. We have to find the Y node next. To find the Y node,

what’s going on with my computer? Oh, there we go. We have the Z node. To find the Y node,

really, we’re just going to take the child of the 65 node that has the taller subtree. So

there’s only one subtree, you know, on the left, there’s no subtree on the right of the 65 node.

So the choice is obvious. But just so you know, if there was, you know, another, you know, child

another you know child hanging off of the right for some reason we would still choose the number

48 node as the y node because it has the taller subtree notice how the subtree on the left is two

and the subtree on the right has a height of one so always take the taller subtree when you’re

looking for uh y and then x so then we do the same thing we go down to the grandchild of the z node

and if we had a choice to go left or right again we would always choose the taller subtree in this

taller subtree in this case there’s only one choice so that x node is going to be the 55

now in this next slide i’ve kind of redrawn this in a little bit different of a way

notice the trinode subtree that we just selected here the 65 48 55 subtree it has a height of three

i mean that should be obvious right it’s just got a height of three

or height. Notice on the left, these are the same three nodes. Exactly. 48, 55, and 65. That’s the

same three nodes that we originally had. But notice how the height is too. So the basic idea for ABL

trees is, you know, compute the balance factors. And as soon as you identify a node that’s out of

whack, you grab a tri-node sub-tree starting at the node that’s out of whack, the Z node.

And then we’ll perform, quote unquote, a rotation where we just kind of smoosh it to be a little

be a little bit more balanced and a little bit more a little bit shorter so the height goes from

three to two once we do that the rest of the tree it’s going to get a little bit better you’ll notice

that this will end up sometimes completely but sometimes partially fixing bad balance factors

in the tree so um this is the end of the intro i only wanted to spend a little time just kind of

like warming you up to the idea in the next videos uh we’re going to actually look at performing

identifying the four different types of rotations that you could see and like how to do them.

We’ll talk more about the idea that we’re really just rearranging pointers here.

If you’re thinking about code, you’re thinking about, oh, I’ve got some pointers between these

nodes. We’re not going to delete any nodes or add any nodes. We’re just going to like

rearrange who is a parent of who, who is a child of who. And then eventually we’ll deal with some

really bad trees and we’ll become experts at just like maintaining ABL trees.

talk about ways to visualize the rotations, which makes it a little bit easier to remember these

patterns and stuff. Okay, so thank you for watching this video. I hope you learned a little bit of

stuff and had a little bit of fun. In the next one, we’re going to get started. We’re going to

go deep. I’m going to get my gear. Sorry.

Hey, everybody. Thanks for watching this video again from the bottom of my heart. I really

from the bottom of my heart. I really appreciate it. I do hope you did learn something and have

some fun. If you could do me a please, a small little favor, could you please subscribe and

follow this channel or these videos or whatever it is you do on the current social media website

that you’re looking at right now? It would really mean the world to me and it’ll help make more

videos and grow this community. So we’ll be able to do more videos, longer videos, better videos,

Do me a kindness and subscribe.

You know, sometimes I’m sleeping in the middle of the night and I just wake up because I know somebody subscribed or followed.

It just wakes me up and I get filled with joy.

That’s exactly what happens every single time.

So you could do it as a nice favor to me or you could troll me if you want to just wake me up in the middle of the night.

Just subscribe and then I’ll just wake up.

I promise that’s what will happen.

Also, if you look at the middle of the screen right now, you should see a QR code which you can scan in order to go to the website.

in order to go to the website which i think is also named somewhere at the bottom of this video

and it’ll take you to my main website where you can just kind of like see all the videos i published

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if you have a suggestion for uh uh clarifications or errata or just future videos that you want to

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The post AVL Tree Tutorial: Balance Factors and Why They Fix Slow BSTs appeared first on NeuralLantern.com.

Clear step-by-step explanation of what defines a Binary Search Tree (BST). We build the definition rule-by-rule starting from graphs all the way to the BST ordering property (all left descendants, node, all right descendants). Great first video before learning insert, delete, search, and Big-O.

What is a Binary Search Tree? 00:00
Intro to Graphs and Nodes 00:42
Connected Graph Requirement 03:16
Acyclic Graph – Removing Cycles 04:54
Turning Graph into Tree 07:05
Establishing a Root 07:09
Adding Hierarchy and Levels 09:05
Single Common Ancestor 11:18
Binary Tree Definition 17:30
BST Ordering Property 17:39
Left Subtree Less Than Node 18:29
Fixing Invalid BST Example 20:26
Valid BST Final Check 21:24
In-Order Ascending Order 24:11
Video Summary and Next Steps 25:02
Thanks and Call to Action 25:30

=-=-=-=-=-=-=-=-=

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hey there let’s talk about binary search trees specifically in this video I want to help you

define and identify what is a binary search tree all the rules that go into determining whether

or not something is a binary search tree so you can you can tell that you’re looking at one or

whether you’re not looking at one we’ll talk about some terminology but keep in mind I’m

going to save searching and sorting and like you know deleting and adding items and a whole bunch

extra stuff, especially the hard stuff for videos that’ll come right after this one.

For now, we’re just going to talk about what is a binary search tree.

So you can see on the screen right here, I hope that we have a binary search tree.

Let’s see.

Can you see that?

Yeah, I think you probably can.

Yeah, I think you can.

This is not drawn quite as well as I would like to draw ours.

I’ll tell you how to do that in a little while.

to do that in a little while but when you draw a tree in a really really nice pretty aligned way

it’s way easier to debug it’s way easier to tell that you’re looking at something that is sorted

and so forth anyway okay so first let’s start off with a graph so this is not a video about graphs

uh i guess i can probably expect that you already kind of know what a graph is at this point in time

and edges. So what do I mean by nodes? Well just imagine there’s like a little circle here

that contains some sort of data. If you know data structures already then

there would be a t-type sitting inside the node. We’ll just call the t-type an integer

and so for our nodes, for our trees, we’re just going to say that they hold integers even though

you know technically speaking in a binary search tree or a graph you could have all sorts of

Okay. So I’m just going to, you know, grab this and sort of like duplicate it and just duplicate it again.

I’m just going to make a bunch of duplicates. Something is definitely wonky about my setup right now.

I think I just upgraded the camera and the CPU is maxed out. I just upgraded the CPU on this computer too.

Oh well. Okay. So we aren’t going to support duplicate values in our binary search tree,

I’m going to say it doesn’t really matter for the graph. Later on, we’ll start making sure there’s

no duplicate data. For now, I think I can probably do that somewhat quickly. Let’s see. So a graph,

you know, for the for our purposes right now, a graph is just basically a collection of nodes and

edges. It could be empty. It could be just nodes. It couldn’t just be edges, it would have an edge

always has to have a node or two connected to it. So we just have like a bunch of nodes inside of

data values this is a valid graph but I’m going to draw some edges randomly here so we’re going

to do this it doesn’t really matter I’m just randomly drawing edges okay actually I should

probably do maybe like one that’s kind of separated okay so this is a graph this is not a binary search

tree so let’s add a bunch of rules let me just pull up my rules real fast here okay so what is

First we’ll start off with the graph like we have here and then we’ll say that the graph must be connected.

What is a connected graph? A connected graph is basically a graph where every single node can find every single other node

or find a path to every single other node only following along edges and only by respecting the direction of edges.

You can see this particular graph is undirected meaning the edges don’t really have a direction to them.

So that means we could travel along in any direction from node to node.

So just as a real quick, you know, I guess like tutorial, I’m going to say is 8 connected to every single other node?

Well, it can find a path to 12 and it can find a path to 33 and it can find a path to 76 and 45.

But it cannot find a path to 99, which means this is not a connected graph.

So that’s the first rule that we have to implement here.

We’re going to say this graph needs to be upgraded or modified so that it will be a connected graph.

start randomly adding edges until this is a connected graph I’m gonna do maybe

like an edge from 99 to 12 there and then now I think this is a connected

graph but let’s double check real fast so connected graph you know what I’m

gonna put graph connectedness in another video to keep this one short for now

check out my other videos if you would like to see whether or not a graph is

but long story short as I just said if every single node can find a path to

every single other node without you know skipping where there is no edge then

it’s connected okay so that’s the first thing the next thing that we need let me

just put the words connected up here we need a connected graph the next thing

that we need is we need an acyclic graph which basically means a graph with no

A cycle is, can you find a path in this graph that starts at one node and ends at the same node without repeating any edges?

So for example, if I was asking, you know, is 76 involved in any cycles?

Well, we could go from 76 to 33 and then 12.

And, you know, we could kind of take 8, 45, 12 again.

But by the time we turn around and try to get back to the 33, we’ve already crossed this edge twice.

so that means the 76 is not involved in any cycles because we cannot find our way back to 76

without repeating an edge so 76 is okay i’m just going to maybe put like a little check box here

and uh you can also tell that the 33 is not involved in any cycles because we go down to

the 76 and then come back up to 33 we already repeated an edge so that’s not valid same thing

again. Same thing for the 34, same thing for the 99. But if you look at the 12 here, we could

actually find a cycle. We could go from the 12 to the 8, and then from the 8 to the 45, and from the

45 to the 12. Now we’re back where we started. We’re at the 12 node again, and we did not repeat

any edges. So this is not a cyclic. This is a cyclic graph. It’s got a cycle. So that’s the

we want to make sure that our graph has no cycles. So, uh, the next rule I’m just going to say is,

is going to be satisfied by me removing some edges at random. So I don’t know, let’s get rid of, uh,

the eight to the 45 connection. See if that works. Um, let’s see. Well, wait a minute. What did I

just do? Something happened here. I just accidentally, oh, I erased too many edges

from the last slide. Okay. Let me add that one back in 33 to 76. That was supposed to be there.

33 to 76 that was supposed to be there.

So now we have an acyclic graph

and we can consider this graph a tree.

The next thing we need to do is add a hierarchy to this tree.

So this is gonna make more sense in a little bit

but for now I’m gonna say,

we’ll imagine a family tree kind of,

it’s not really exactly a family tree

but you know imagine there are parents

and children and grandchildren

and we’ll just try to like,

we’ll try to rank children according to

according to you know when they were born or or how many parents or children

they have you know their descendancy their lineage whatever you want to call

it so I think I’m just gonna maybe move the eight up a little bit so that it’s

on the same rank with the 12 and the 33 so I’m just gonna move these up a little

bit and then redo their connecting lines so I’m gonna do this and that and then

and then we’ll say that the 76 is still a child of the 33.

So I’ll just kind of move it over here.

Or sorry, we never said it was a child.

Now we’re saying it’s a child

because when you go from top to bottom,

imagine like a family tree,

we’ll say we have parents and children.

The children or the descendants are lower.

The ancestors or the parents or the grandparents are higher.

Maybe I want to put this 45 over on the left

so that it’s a child of the 8 node.

And maybe I’ll say that the 99 node

the 99 node is a child of the 12 node. Okay, so eventually this is going to look not exactly like

a family tree because usually in the family tree we have two parents that go to one or more children.

In this type of tree, in eventually our binary tree, we’re going to have one parent that has,

you know, children just by itself, like asexual reproduction. We’re pretending to be bacteria

What was that comedy skit a long time ago, Tiny Elvis?

And he was like, hey, see that protozoa over there?

The thing is huge.

Anyone?

No?

Okay.

So the next rule we need is we must have a rooted tree.

What is a rooted tree?

Actually, let me clean up the lineage real fast, the heritage.

I want to make nodes that have the same level with respect to their ancestors

respect to their ancestors to have the same physical y coordinate the same

physical level so notice how the 99 I just moved it down a little bit because

I wanted it to look like it’s the same you know number of generations lower

than the top row so like with aid it had a child 45 so that’s one level down 33

had a child that’s one level down that’s a 76 and the 12 also one level down

would have been 99 but the 99 was kind of too physically high this will help

will help you debug your binary trees in the future anyway so i’m just going to clean this up right now

and uh now that we have you know parent child relationships we need to look at the most common

ancestor in the whole entire tree or if there is one so notice how uh i guess i’m just kind of using

the word ancestor without explaining it so your ancestors are the people that you’re disappointing

or whatever it is, ancestors came before us, right?

Like my parents are my ancestors.

My grandparents are also my ancestors.

My great-grandparents are also my ancestors.

Going down in the family tree are your descendants.

You know, my children are my descendants.

My grandchildren are my descendants.

My great-grandchildren are my descendants and so forth.

So if you look at the number 34 here,

what is the greatest ancestor that it has?

Well, if you just go up and up and up

until you can’t go up anymore,

up until you can’t go up anymore it’s the 12th so that means the greatest ancestor of at least in

this tree of the 34 is the 12th the 45 the highest you can go is the 8 notice how 8 and 12 and 33 are

siblings there’s not really like a parent-child relationship they’re connected sideways we’ll

eventually get rid of the sideways connections for our binary search trees but for now we have

some siblings and so we actually have three different greatest ancestors in this tree we’ve

tree we’ve got 8 and 12 and 33 and that’s no good so the next rule that we need is we need

one common ancestor for every single other node in the entire tree so I’m going to duplicate this

slide right here and I’m going to say we need to clean this up so we got to choose 8 or 12 or 33

I don’t know I’m just going to maybe say that 8 and it’s 45 are children of the 12 so I’m just

line here arbitrarily so i’m gonna go like that notice how i’m trying to keep the 8 the 99 and

the 76 on the same y coordinate the same level now we have two uh you know greatest ancestors

we need to have only one so i think i’m probably going to say that the 33 and the 76 are going to

become children of the 12 node just for simplicity so i’m going to put this over here i’m going to

i’m gonna like move this over here this is a gross tree but we’ll fix it soon we’ll do that all right

okay okay

okay and then i have to get rid of that little sibling line and then i have to add another line

to say that 33 is a child of 12. okay so now we have a rooted tree what we had before was an

unrooted tree or just like not a rooted tree because there was no root there was no greatest

greatest common ancestor to the entire rest of the tree. But now we can say that 12 is the root

of the tree because it’s the ancestor for all other nodes. I guess even if 12 was by itself,

we could call it a root of tree. It just wouldn’t be a very good example. So I’m going to actually

get rid of the word acyclic here since we already handled that. And I’m going to do,

I’m going to say that 34’s greatest ancestor is 12, 99’s greatest ancestor is 12, 33’s greatest

all the nodes in the tree their greatest ancestor is 12 and it’s common to all of them so now we

have a rooted tree maybe I should type in my doomed I really hope the screen I really hope

the screen cast doesn’t freeze it probably will upgraded the CPU it’s 33 faster than it was before

board so i am still using an old cpu even though it’s new to me dang it i’m gonna have to spend

like 500 okay uh so the next rule is so we have a rooted tree and now let’s rearrange everything

hierarchically which we already kind of did so uh there’s not really much to do here except

we did hierarchical organization

we probably just need to move on to our next rule which is to say that

if our rooted tree is to become a binary tree

then each node in the entire tree

cannot have more than two children

every single node in the entire tree

can have zero or one or two children

if there are three or more anywhere in the entire tree

then it’s not a binary tree

so okay that means you know the 45 is okay

because it has no children

because it has no children. All the rows, all the leaves at the bottom are okay. All these nodes

over here are okay because they just have one child. But the 12 node, the root of the tree,

it’s got three children and that violates the rule. So now we need to rearrange this.

So I think I’m going to just move these numbers down here maybe. Whoops. Maybe move these nodes

descendants of the 33 node. So I’m going to do this.

Okay. So cool. Now we have a binary tree and now we need to rearrange the drawing a little bit.

This is not part of the definition, but this will make it easier to debug your trees. So what I want

to do is make it so that every left descendant of a node appears physically on that node’s left

every right descendant of a node should appear on its right side

or like further to the right on the x-coordinate

so sort of similar to you know like the y-coordinate ranking

but notice how the 12 its left child is 8 and its right child is 33

so that means all the nodes here on the left side you know that come underneath the 8

are left descendants or you could say that they are in the left subtree of the 12.

So, they’re okay because 8 is to the left of 12 and 45 is to the left of 12.

But notice how 33 is a right child of 12 and it’s part of the right subtree.

It’s part of the right descendancy, but it’s physically on the left of 12.

That’s going to be really bad because it’ll be harder to debug our trees later.

So, I’m just going to slide it over without rearranging the tree.

And I usually try to split the difference here.

and then their parent is kind of like physically in the middle that will also help you debug in

the future so i’m going to do this wrong let me get the blue i’m going to do this make sure that

my ranking is still okay not the best most neatest tree and so now i’m going to look at the eight

it’s left subtree or it’s left uh descendancy uh all of those nodes which is just the 45 are

it’s left child 76 is the only thing that’s on the left side in the left subtree so that’s all

right probably the 33 I think it’s still a little lopsided I probably could have you know dragged

it a little bit more to the right but I’m going to leave it so the 33 and the 76 are okay and then

the 99 is a right descendant or it’s part of the right subtree of the 33 so it should be on the

right side it is that’s good the 34 should be on the right side of the 33 because it’s it’s a it’s

It’s part of the right subtree of 33.

It’s part of the right descendancy.

And it should also be on the left side of the 99

because it’s part of the left subtree of the 99.

So, so far, this is okay.

This is not the best drawing.

So now we have a binary tree

and we have kind of drawn it in a slightly good way.

Not super great.

The last rule that we’re going to add here is,

whoops, what happened there?

I’m supposed to do that, yeah.

The last rule we’re going to add is

we must have ordering to our nodes, to the numbers in our nodes. So what do I mean by ordering? Well,

the whole reason that we want a binary search tree, at least usually, is because it’s really,

really fast to search through. This is going to be a tree that we can search through in log time,

which we’ll talk more about in another video. We’ll do time complexities and searching and stuff. But

in order for this to be a super fast tree to search through, we need to be able to

very quickly make a decision. Do we go down and to the left? Do we go down and to the right?

left do we go down into the right and every decision we make should help us eliminate half

of the remaining data set half of the tree at a time every time we go down a level and that’s only

possible if the nodes themselves are ordered so what i’m going to do is say every node that’s on

the left side or let’s say every descendant that’s on the left side of a node should be of a lesser

value than that particular node every node that is on the right side of a node meaning it’s a right

descendant of that node or part of the right subtree of the node should be greater than that

node. Please note that this would not allow us to support duplicate values in our tree.

You could use something like less than or equal to versus greater than if you wanted to support

duplicate values in your tree. You could also change the T type, you know, the template type

to a node from the data you want it to store to a list. So every single node has a list inside of it

and the items in the list are just, you know, the regular values that you wanted. There’s a whole

values that you wanted there’s a whole bunch of different ways you could actually do that

but for now we’re going to see our trees don’t support duplicate values

so okay we don’t support duplicate values but we need you know from left to right

the order needs to look sane and this is kind of why we’re drawing the tree in this special way

we’re trying to make our tree really easy to debug visually so check this out the way we’ve

drawn this every single node has its own x coordinate basically notice how the 12 there’s

nothing underneath it at least not directly underneath it the 33 the 34 all the nodes you

know you can draw a straight line down from the node and it shouldn’t cross over with any other

nodes or edges so that means now that we’ve drawn it this way and we’ve made sure that all the left

descendants are on the left and all the right descendants are on the right we could actually

just scan our eyes from left to right and be sure that we are seeing an increasing order or

Since duplicates aren’t allowed in this tree, we’ll say an increasing order or an ascending order.

So if we start at 45 and then we go to the right, we should see a higher number.

So if we look at 8, 8 is a lower number.

So this violates the ordering rule.

This is not a valid binary search tree.

It’s only a valid binary tree.

So sad.

Okay, so I could, you know, swap the 45 and the 8 or I could just increase the value of the 8 itself.

of the 8 itself, I think it’ll be faster for me to just increase the 8 value. So I’m going to make

it 52 now. So we’ve got 45, 52. Then we go to the right a little bit more. It should also increase.

So 52 should increase. I’m just going to randomly increase it to, you know, 59, let’s say. Then as

we go to the right, we see another number here. It is 76. That already is increased from 59.

So we don’t need, we don’t need to alter that number. Then we go to the right one more time

that because it’s lower than 76 so I’m going to put 87 then when we go to the right one more time

that’s the 34 down here that’s definitely a decrease so we have to increase it now so I’m

going to say 99 how about if I just do 92 yeah so now we have a 92 down here and then if we go

to the right a little bit more there’s the 99 and now we actually have a binary search tree let’s

just double check all the rules first off is this a graph yeah there’s nodes and edges there’s nothing

There’s nodes and edges.

There’s nothing weird happening.

There’s not like a edge floating in the middle of nowhere.

Is this a connected graph?

Yeah, it is.

Because every single node, if you followed, you know, the edges, you know, could find

every single other node.

Like the 45, it could find, you know, the 92 node if it just followed all these edges.

Notice again that these edges have no direction.

There’s no arrows on them.

So we could go up or down in any direction we want.

So that means the 76 can find everything else.

else and so this is a connected graph for sure okay now that we have a

connected graph is this an acyclic graph I don’t know about you but I can’t find

any cycles in the graph right like the 92 it goes out but it doesn’t come back in

same thing for the 76 if we went from the 76 up to the 87 and then took a

right turn we wouldn’t be able to come back towards it without repeating that

same edge so this is definitely an acyclic graph all of the nodes follow

they’re not involved in cycles is this a rooted tree yeah there’s only one common ancestor to the

entire tree it’s the 59 it’s the highest node in the entire tree and it’s common to all other of

all of its descendants um so let’s see i’m going to duplicate this real fast the next thing is

we arranged it hierarchically and we added the terminology of like i think i kind of just like

tried to sneak this one past you but we added the terminology of a left child and a right child

and a right child.

Right? We also added the terminology of a right subtree and a left subtree.

Just to clarify what I mean, what do I mean by a right subtree and a left subtree.

So imagine we’re considering the 59.

If we consider the 59, then the 52 is the root node of the left subtree of the 59 node.

If we consider the 59, then the 87 is the root node of the right subtree of the 59 node.

so the right child is basically the root of the subtree uh in in you know the right child is the

root of a subtree of the right subtree uh i think i’m getting muddled my words here okay

anyway so we have a little bit of terminology uh the 59 is the root node that’s like the

the only node in the entire tree without uh an ancestor or a parent so okay uh then we have a

Note, and the entire tree doesn’t have more than two children.

Notice how the 59 has two children.

That’s fine.

If it was three, that’s way too much.

52 has one.

87 has one.

No, 87 has two children.

99 has one child.

92 has zero children.

So, okay, we’re not violating that rule.

So, we definitely have a binary tree.

And then we, you know, kind of read from left to right.

Let me write down the numbers this time for fun.

physically without kind of like following edges or anything we just scan

our eyes from left to right we can see that we have an ascending list if we

supported duplicates it would be okay to just see a non decreasing list but we

see an increasing list or an ascended list ascending list okay whoops this is

left to right so this is actually a binary search tree and um i don’t know about you but i think i

need to like you know get a snickers or something right now because uh this video has been going on

for longer than i wanted it to go on so we’ve defined a binary search tree in future videos

we’ll talk about how to add nodes into a binary search tree how to build the tree what’s the time

complexity of the tree in various operations deleting nodes from a tree terminology like

terminology like what are these nodes called things like that so anyway thank you so much

for for watching i hope you learned a little bit of stuff and had a little bit of fun

i’ll see you at a later date in time

just kidding

hey everybody thanks for watching this video again from the bottom of my heart i really

appreciate it i do hope you did learn something and have some fun uh if you could do me a please

If you could do me a please a small little favor

Could you please subscribe and follow this channel or these videos or whatever it is you do on the current social media?

It’s a website that you’re looking at right now

It would really mean the world to me and it’ll help make more videos and grow this community

So we’ll be able to do more videos longer videos better videos or just i’ll be able to keep making videos in general

So please do do me a kindness and uh and subscribe

I’m sleeping in the middle of the night and I just wake up because I know somebody subscribed or followed.

It just wakes me up and I get filled with joy.

That’s exactly what happens every single time.

So you could do it as a nice favor to me or you could troll me if you want to just wake me up in the middle of the night.

Just subscribe and then I’ll just wake up.

I promise that’s what will happen.

Also, if you look at the middle of the screen right now, you should see a QR code which you can scan in order to go to the website,

which I think is also named somewhere at the bottom of this video.

and it’ll take you to my main website where you can just kind of like see all the videos

I published and the services and tutorials and things that I offer and all that good

stuff.

And if you have a suggestion for clarifications or errata or just future videos that you want

to see, please leave a comment.

Or if you just want to say, Hey, what’s up?

What’s going on?

You know, just send me a comment, whatever.

I also wake up for those in the middle of the night.

I get, I wake up in a cold sweat and I’m like, it would really, it would really, it

it would really mean the world to me.

I would really appreciate it.

So again, thank you so much for watching this video

and enjoy the cool music

as I fade into the darkness,

which is coming for us all.

Thank you.

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