AVL Trees 2.pdf

Degenerate Binary Search Tree

BST for 14, 15, 4, 9, 7, 18, 3, 5, 16, 20, 17

14

15 4

9

7

18 3

5

16 20

Degenerate Binary Search Tree

BST for 3 4 5 7 9 14 15 16 17 18 20

14

15 4

9 7

18 3

5

16

Degenerate Binary Search Tree

BST for 3 4 5 7 9 14 15 16 17 18 20

Balanced BST

—

 

We should keep the tree balanced.

—

 

One idea would be to have the left and

Balanced BST

Does not force the tree to be shallow.

14

15

18 16

20 17

4

9 7

3

Balanced BST

—

 

We could insist that every node must have left

and right subtrees of same height.

—

 

But this requires that the tree be a complete

binary tree

—

 

To do this, there must have (2d+1 – 1) data

AVL Tree

—

 

AVL (Adelson-Velskii and Landis) tree.

—

 

Purpose: AVL is a self-balancing binary tree

—

 

This was an effort to propose the development

of a balanced search tree by considering the height as a standard.

—

 

An AVL tree is identical to a BST except

§

 

height of the left and right subtrees can differ by

at most 1.

AVL Tree

—  An AVL Tree

5

8 2

4

3

1 ₇

level 0

1

2

AVL Tree

—

 

Not an AVL tree

6

8 1

4

3 1

5

level 0

1

2

Balanced Binary Tree

—

 

The height of a binary tree is the maximum level

of its leaves (also called the depth).

—

 

The balance of a node in a binary tree is defined

as the height of its left subtree minus height of its right subtree.

—

 

For example, a tree is a balanced tree if each

Balanced Binary Tree

-1

0 1

0 0

0

0

-1 0

1

0

0 0

0 0

0 0

Balanced Binary Tree

Insertions and effect on balance

-1 0 1 0 0 0 0 -1 0 1 0 0 0 0 0 0 0

U₁ U₂ U₃ U₄

U₅ U₆ U₇ U_{8 U}

9 U10 U11 U12

B B

B B

Balanced Binary Tree

—

 

Tree becomes unbalanced only if the newly

inserted node

§

 

is a left descendant of a node that previously

had a balance of 1 (U₁ to U₈),

§

 

or is a descendant of a node that previously had

Balanced Binary Tree

Insertions and effect on balance

-1 0 1 0 0 0 0 -1 0 1 0 0 0 0 0 0 0

U₁ U₂ U₃ U₄

U₅ U₆ U₇ U_{8 U}

9 U10 U11 U12

B B

B B

Balanced Binary Tree

Consider the case of node that was previously 1

-1 0 1 0 0 0 0 -1 0 1 0 0 0 0 0 0 0

U₁ U₂ U₃ U₄

U₅ U₆ U₇ U_{8 U}

9 U10 U11 U12

B B

B B

Inserting New Node in AVL Tree

1

0

A

B

T₁

T₃

Inserting New Node in AVL Tree

2

1

A

B

T₁

T₃

T₂

new

1

Inserting New Node in AVL Tree

2 1 A B T₁ T₃ T₂ new 1 2 0 0 A B T₁ T₃ T₂ new

AVL Tree Building Example

—

 

Let us work through an example that inserts

numbers in a balanced search tree.

—

 

We will check the balance after each insert

AVL Tree Building Example

Insert(1)

AVL Tree Building Example

Insert(2)

1

AVL Tree Building Example

Insert(3) single left rotation

1

2

3

AVL Tree Building Example

Insert(3) single left rotation

1

2

3

AVL Tree Building Example

Insert(3)

1

2

26

AVL Tree Building Example

Insert(4)

1

2

3

27

AVL Tree Building Example

Insert(5)

1

2

3

4

5

28

AVL Tree Building Example

Insert(5)

1

2

3

4

29

AVL Tree Building Example

Insert(6)

1

2

3

4

5

6

30

AVL Tree Building Example

Insert(6)

1

2

3 4

5

31

AVL Tree Building Example

Insert(7)

1

2

3 4

5

6

7

32

AVL Tree Building Example

Insert(7)

1

2

3 4

5

6

33

AVL Tree Building Example

1

2

3 4

5

6

7

34

AVL Tree Building Example

Insert(15)

1

2

3 4

5

6

7

16

15

35

AVL Tree Building Example

Insert(15)

1

2

3 4

5

6

7 16

15

36

Cases for Rotation

—

 

Single rotation does not seem to restore the

balance.

—

 

The problem is the node 15 is in an inner

subtree that is too deep.

37

Cases for Rotation

—

 

Let us call the node that must be rebalanced α .

—

 

Since any node has at most two children, and a

height imbalance requires that α’s two subtrees

differ by 2 (or –2), the violation will occur in four cases:

1.

 

An insertion into left subtree of the left child of α

2.

 

An insertion into right subtree of the left child of α

3.

 

An insertion into left subtree of the right child of α

4.

 

An insertion into right subtree of the right child

38

Cases for Rotation

Outside insertion:

—

 

The insertion occurs on the “outside” (i.e.,

left-left or right-right) in cases 1 and 4

—

 

Single rotation can fix the balance in cases 1

and 4.

Inside insertion:

—

 

Insertion occurs on the “inside” in cases 2

39

Cases for Rotation

Single right rotation to fix case 1. (LL)

—

 

How to add a new node as a child of node X.

40

Cases for Rotation

Single left rotation to fix case 4. (RR)

—

 

Similarly, How to add a new node as a child of

node X.

Level n Level n-1 Level n-2

k₁

k₂ X

Y

Z

k₁

k₂

41

Cases for Rotation

Single right rotation fails to fix case 2. (RL)

—

 

How about if the new node a child of node Y

k₁ k₂ Z X Level n Level n-1 Level n-2 k₁ k₂ Z X

α α

new

Y

new

42

Cases for Rotation

—

 

Y is non-empty because

the new node was inserted in Y.

—

 

Thus Y has a root and two

subtrees.

—

 

View the entire tree with

four subtrees connected with 3 nodes.

k₁

k₂

X

Y

43

Cases for Rotation

—

 

Y is non-empty because

the new node was inserted in Y.

—

 

Thus Y has a root and two

subtrees.

—

 

View the entire tree with

four subtrees connected with 3 nodes.

k₁

k₃

D

A

B C

44

Cases for Rotation

k₃ k₁ A new 1 2

New node inserted a either of the two spots

B C

k₂

D

new’

§  Exactly one of tree B or C

is two levels deeper than

D; we are not sure which

one.

§  Good thing: it does not

matter.

45

Cases for Rotation

k₃ k₁ A new 1 2

New node inserted a either of the two spots

B C

k₂

D

new’

§

 

A rotation between

k

₃

and

k

₁

(

k

₃

was

k

₂

then)

was shown to not work.

§

 

The only alternative is

to place

k

₂

as the new

root.

§

 

This forces

k

₁

to be

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Cases for Rotation

—

 

Left-right

double

rotation to fix case 2.

New node inserted a either of the two spots

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Cases for Rotation

—

 

Left-right

double

rotation to fix case 2.

A new B D C new’ A new B D C new’ k₃ k₂ k₁

Rotate right k2

48

Cases for Rotation

—

 

Right-left

double

rotation to fix case 3 (LR)

k₁

k₃

D A

B C

k₂

k₁

k₃

D A

B

C k₂

49

Cases for Rotation

—

 

Right-left

double

rotation to fix case 3.

k₁

k₂

D

A B C

k₃ k₁

k₃

D A

B

C k₂

50

AVL Tree Building Example

Insert(15)

1

2

3 4

5

6

X


(null)

Z


(null) Y

15

k₁

k₂

7

51

AVL Tree Building Example

Insert(15) right-left double rotation

1

2

3 4

5

6

7

16

15

k₁

k₃

k₂

52

AVL Tree Building Example

Insert(15) right-left double rotation

1

2

3 4

5

6

7

16 15

k₁

k₃ k₂

53

AVL Tree Building Example

Insert(15) right-left double rotation

1

2

3 4

5

6

16

k₃

15

k₂

7

54

AVL Tree Building Example

Insert(14): right-left double rotation

1

2

3 4

5

6

16

k₃

15

k₂

7

k₁

14

55

AVL Tree Building Example

Insert(14): right-left double rotation

1

2

3 4

5

6

16

k₃

15

k₂

7

k₁

14

56

AVL Tree Building Example

Insert(14): right-left double rotation

1

2

3 4

5 ₁₆

k₃

15

k₂

7

6

k₁

57

AVL Tree Building Example

Insert(13): single rotation

1

2

3 4

5 ₁₆

15 7

6

14

13

58

AVL Tree Building Example

Insert(13): single rotation

1

2

3 4

5

16 15

7

6 14

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Deletion in AVL Tree

—

 

Delete is the inverse of insert: given a value X

and an AVL tree T, delete the node containing X and rebalance the tree, if necessary.

—

 

Turns out that deletion of a node is

60

Deletion in AVL Tree

—

 

Insertion in a height-balanced tree requires at

most one single rotation or one double rotation.

—

 

We can use rotations to restore the balance

when we do a deletion.

—

 

We may have to do a rotation at every level of

61

Deletion in AVL Tree

—

 

Here is a tree that causes this worse case

number of rotations when we delete A. At every node in N’s left subtree, the left subtree is one

shorter than the right subtree.

A C

D

N

E J

G

I F

H

K L

62

Deletion in AVL Tree

—

 

Deleting A upsets balance at C. When rotate (D

up, C down) to fix this

A C

D

N

E J

G

I F

H

K L

63

Deletion in AVL Tree

—

 

Deleting A upsets balance at C. When rotate (D

up, C down) to fix this

C

D

N

E J

G

I F

H

K L

64

Deletion in AVL Tree

—  The whole of F’s left subtree gets shorter. We fix this

by rotation about F-I: F down, I up.

C

D

N

E

J G

I F

H

K L

65

Deletion in AVL Tree

—  The whole of F’s left subtree gets shorter. We fix this by

rotation about F-I: F down, I up.

C

D

N

E

J G

I F

H

K L

66

Deletion in AVL Tree

—

 

This could cause imbalance at N.

—

 

The rotations propagated to the root.

C

D

N

E

J G

I

F

H

K

L

67

Deletion in AVL Tree

Procedure

—

 

Delete the node as in binary search tree (BST).

—

 

The node deleted will be either a leaf or have

just one subtree.

—

 

Since this is an AVL tree, if the deleted node has

one subtree, that subtree contains only one node (why?)

—

 

Traverse up the tree from the deleted node

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Deletion in AVL Tree

—

 

There are 5 cases to consider.

—

 

Let us go through the cases graphically and

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Deletion in AVL Tree

Case 1a: The parent of the deleted node had a balance of 0 and the node was deleted in the parent’s left subtree.

Action: change the balance of the parent node and

stop. No further effect on balance of any higher node.

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Deletion in AVL Tree

Here is why; the height of left tree does not change.

How about if we delete(1). The parent is 2 with balance of 0.

1

2

3 4

5

6

7

0

1

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Deletion in AVL Tree

1 2 3 4 5 6 7 2 3 4 5 6 7 0 1 2 remove(1)

Here is why; the height of left tree does not change.

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Deletion in AVL Tree

Case 1b: the parent of the deleted node had a balance of 0 and the node was deleted in the parent’s right subtree.

Action: (same as 1a) change the balance of the

parent node and stop. No further effect on balance of any higher node.

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Deletion in AVL Tree

Here is why; the height of left tree does not change.

How about if we delete(7). The parent is 2 with balance of 0.

1

2

3 4

5

6

7

0

1

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Deletion in AVL Tree

1 2 3 4 5 6 7 2 3 4 5 6 0 1 2 Remove(7)

Here is why; the height of left tree does not change.

How about if we delete(7). The parent is 2 with balance of 0.

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Deletion in AVL Tree

Case 2b: the parent of the deleted node had a balance of –1 and the node was deleted in the parent’s right subtree.

Action: same as 2a: change the balance of the parent

node. May have caused imbalance in higher nodes so continue up the tree.

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Deletion in AVL Tree

Case 3a: the parent had balance of –1 and the node was deleted in the parent’s left subtree,

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Deletion in AVL Tree

Case 3a: the parent had balance of -1 and the node was deleted in the parent’s left subtree,

right subtree was balanced.

Action: perform single rotation, adjust balance. No

effect on balance of higher nodes so stop here.

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Deletion in AVL Tree

Case 4a: parent had balance of –1 and the node was deleted in the parent’s left subtree, right

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Deletion in AVL Tree

Case 4a: parent had balance of -1 and the node was deleted in the parent’s left subtree, right

subtree was unbalanced.

Action: Double rotation at B. May have effected the balance of higher nodes, so continue up the tree.

80

Deletion in AVL Tree

Case 5a: parent had balance of –1 and the node was deleted in the parent’s left subtree, right

81

Deletion in AVL Tree

Case 5a: parent had balance of -1 and the node was deleted in the parent’s left subtree, right

subtree was unbalanced.

Action: Single rotation at B. May have effected the

balance of higher nodes, so continue up the tree. rotate