avl-trees: motivation

1

AVL-Trees: Motivation• Recall our discussion on BSTs

– The height of a BST depends on the order of insertion• E.g., Insert keys 1, 2, 3, 4, 5, 6, 7 into an empty BST• Problem: Lack of “balance” – Tree becomes highly

asymmetric and degenerates to a linked-list!!

– Since all operations take O(h) time, where log N <= h <= N-1, worst case running time of BST operations are O(N)

• Question: Can we make sure that regardless of the order of key insertion, the height of the BST is log(n)? In other words, can we keep the BST balanced?

2

Height-Balanced Trees• Many efficient algorithms exist for

balancing BSTs in order to achieve faster running times for the BST operations– Adelson-Velskii and Landis (AVL) trees (1962)– Splay trees and other self-adjusting trees (1978)– B-trees and other multiway search trees (1972)– Red-Black trees (1972)

• Also known as Symmetric Binary B-Trees• Will not be covered in this course

3

AVL Trees: Formal Definition1. All empty trees are AVL-trees2. If T is a non-empty binary search tree with

TL and TR as its left and right sub-trees, then T is an AVL tree iff1. TL and TR are AVL trees2. |hL – hR| <= 1, where hL and hR are the heights

of TL and TR respectively

hLhL+1 or hL-1

TL TR

4

AVL Trees• AVL trees are height-balanced

binary search trees

• Balance factor of a node = height(left subtree) - height(right subtree)

• An AVL tree can only have

balance factors of –1, 0, or 1 at every node

• For every node, heights of left and right subtree differ by no more than 1

6

4

1

7

9 0

-1

0

0

1

Red numbersare Balance Factors

An AVL Tree

5

AVL Trees: Examples and Non-Examples

6

5

4

7

9 0

-1

0

0

1

Red numbers are Balance Factors

An AVL Tree

6

4

1

9 0

1

0

0

An AVL Tree

5

6

4

1

9 1

0

0

0

An AVL Tree

5 8 0

6

4

1

9 2

-1

0

1

Non-AVL Tree

7 -1

8 0

6

4

3

7 -2

0

1

2

Non-AVL Tree

8

9 010

-1

6

4

1

10 2

-1

0

1

Non-AVL Tree

8 0

9 070

0 0

6

AVL Trees: Implementation• To implement AVL-trees, associate a

height with each node “x”

left right

key

height

x

• Balance factor (bf) of x = height of left subtree of x – height of right subtree of x

• In an AVL-tree, “bf” can be one of {-1, 0, 1}

C++ Declaration

struct AVLTreeNode{

int key;

int height;

AVLTreeNode left;

AVLTreeNode right;

};

7

Good News about AVL Trees• Can prove: Height of an AVL

tree of N nodes is always O(log N)

• How? Can show:– Height h = 1.44 log(N)– Prove using recurrence

relation for minimum number of nodes S(h) in an AVL tree of height h:• S(h) = S(h-1) + S(h-2) + 1

– Use Fibonacci numbers to get bound on S(h) bound on height h

6

5

4

7

9 0

-1

0

0

1

Red numbersare Balance Factors

An AVL Tree

Height = O(logN)

8

Good and Bad News about AVL Trees• Good News:

– Search takes O(h) = O(logN)

• Bad News• Insert and Delete may cause the tree to be unbalanced!

6

5

4

7

9 0

-1

0

0

1

6

5

4

7

9 0

-1

1

1

2

3

Insert 3

0An AVL Tree

No longer an AVL Tree

9

Restoring Balance in an AVL Tree• Problem: Insert may cause balance factor to

become 2 or –2 for some node on the path from root to insertion point

• Idea: After Inserting the new node1. Back up towards root updating balance factors along the

access path2. If Balance Factor of a node = 2 or –2, adjust the tree by

rotation around deepest such node.6

5

4

7

9 0

-1

1

1

2

30

Non- AVL Tree

10

Restoring Balance: Example

6

5

4

7

9 0

-1

0

0

1

6

5

4

7

9 0

-1

1

1

2

3

Insert 3

0AVLNot AVL

Rotate6

4

3

7

9

-1

0

0

0

AVL

5 0 0

• After Inserting the new node1. Back up towards root updating heights along the access

path2. If Balance Factor of a node = 2 or –2, adjust the tree by

rotation around deepest such node.

11

AVL Tree Insertion (1)

P

L R

A B C D

• Let the node that needs rebalancing be P.– P is called the pivot node– P is the first node that has a bf of 2 or -2 as we

backup towards the root after an insertion

P is the Pivot: bf is 2 or -2

12

AVL Tree Insertion (2)

P

L R

A B C D

• There are 4 cases:– Outside Cases (require single rotation) :

1. Insertion into left subtree of left child of P (LL Imbalance).2. Insertion into right subtree of right child of P (RR

Imbalan.)

− Inside Cases (require double rotation) :3. Insertion into left subtree of right child of P (RL

Imbalance)4. Insertion into right subtree of left child of P (LR

Imbalance)

P is the Pivot: bf is 2 or -2

13

LL Imbalance & Correction

P

L R

A B C D

• LL Imbalance: We have inserted into the left subtree of the left child of P (into subtree A)– bf of P is 2– bf of L is 0 or 1

– Correction: Single rotation to the right around P

2

0 or 1

Tree after insertion

L

R

C D

A

P

B

Tree after LL correction

Rotate

14

LL Imbalance Correction Example (1)

10

4 20

3

1

1

0

0

Initial AVLTree

Insert(2)

• LL Imbalance:– bf of P(4) is 2– bf of L(3) is 0 or

1

10

4 20

3

1

0

0

Tree right after insertion of 2

20

1

Now, backup the tree updating

balance factors

Backup

Classify the type of imbalance

Identified 4 as the pivot

10

4 20

3

1

2

1

0

20

0

Rotate10

4

20

2

1

0

0

03

0

AVL Tree afterLL correction

15

LL Imbalance Correction Example (2)

10

4 20

3

1

0

0

Initial AVLTree

Insert(2) Rotate

5

0

0

0

0

10

4 20

3

1

0

0


2

5 0

0

10

4

20

31

02 50 0

0

AVL Tree afterLL Correction

• LL Imbalance:– bf of P(10) is 2– bf of L(4) is 0 or 1

0

1

10

4 20

3

2

1

0

Identified 10 asthe pivot

2

5 0

Backup

Classify the type of imbalanceNow, backup the

tree updating balance factors

16

RR Imbalance & Correction

P

L R

A B C D

• RR Imbalance: We have inserted into the right subtree of the right child of P (into subtree D)– bf of P is -2– bf of R is 0 or -1

– Correction: Single rotation to the left around P

-2

0 or -1


R

LC

D

A

P

B

Tree after RR correction

Rotate

RR Imbalance Correction Example (1)

10

4 20

30

-1

0

Initial AVLTree

0

-1

Insert(40) Rotate10

4 20

30

-1

0

-1

-2

40 0

Identified 20as the pivot

10

4

20 40

-1

0

0

30

0

0

AVL Tree afterRR Correction

• RR Imbalance:– bf of P(20) is -2– bf of R(30) is 0 or -

1

10

4 20

30

-1

0

0

-1

40 0


Backup



balance factors

RR Imbalance Correction Example (2)

10

4 20

30

-1

0

Initial AVLTree

0

0

Insert(40) &Backup

Rotate


1

150

10

4 20

30

-10

-1

-2

40 0

Tree after insertion of 40

150

10

4 400 0

30

0

-1

AVL Tree afterRR Correction

20

15

0

0

As we backed up the tree, we

updated balance factors and

identified 10 as the pivot


LR Imbalance & Correction

P

L R

B1

2

-1


1

2

B2

LR

Rotate(1)

A C D

Rotate(2)P

L

R

AB1

C D

2

B2

LR

Tree after rotate(1)

PL

R

B1 B2

LR

A

C D

Tree after LR correction

• LR Imbalance: We have inserted into the right subtree of the left child of P (into subtree LR)– bf of P is 2– bf of L is -1

– Correction: Double rotation around L & P

20

LR Imbalance Correction Example

10

4 20

7

1

0

0

0

Initial AVLTree

Insert(5) &Backup

Rotate

• LR Imbalance:– bf of P(10) is

2– bf of L(4) is -1

0 3

AVL Tree afterLR Correction

7

4

3 05

10

020

0

0

0

-1


0

10

4 20-1

2

0

5

73 1

0

1

2





RL Imbalance & Correction

P

L R

C1

-2

2

C2

RL

A DB

1


P

L

R

C1

-2

2

C2

RL

A

D

B

Tree after Rotate(1)

Rotate(1)

P

L

R

C1 C2

RL

A

D

B

Tree after RL correction

Rotate(2)

1

• RL Imbalance: We have inserted into the left subtree of the right child of P (into subtree RL)– bf of P is -2– bf of R is 1

– Correction: Double rotation around L & P

22

RL Imbalance Correction Example

10

4 20

15

-1

0 0

Initial AVLTree

Insert(17)

• RL Imbalance:– bf of P(10) is -2– bf of R(20) is 1

300 0

Rotate 15

20 1

30 017

10

4 0

1

AVL Tree afterRL Correction

0

010

4 20

15

-2

0 1

30-1 0

17 0


2

1





23

Deletion• Deletion is similar to insertion

• First do regular BST deletion keeping track of the nodes on the path to the deleted node

• After the node is deleted, simply backup the tree and update balance factors− If an imbalance is detected, do the appropriate

rotation to restore the AVL tree property− You may have to do more than one rotation as

you backup the tree

24

Deletion Example (1)

10

4 20

3

1

0

0

Initial AVLTree

Delete(20) Backup

5

0

0


Tree after deletion of 20

10

4

3

1

0 5

0

0

Idenfitied 10 as the pivot

10

4

3

2

0 5

0

0

Classify the typeof imbalance


balance factors

Rotate4

10 1

5

3

0

0

-1


25


10

4 20

1

0

Initial AVLTree

Delete(20) Backup

5

-1

0

• LR Imbalance:– bf of P(10) is

2– bf of L(4) is -1


10

4

1

5

-1

0

Idenfitied 10 as the pivot

10

4

2

5

-1

0

Classify the typeof imbalance


balance factors

Rotate5

10 040

0


1

2

26


10

5 20

1

0

Initial AVLTree

Delete(10)

7

0

3 15 30

402 4

1

8

1

1

1 0 0

-1

-1

0 -1

15

5 20

1

0


We have copied the successor of 10 to root

and deleted 15

7

0

3 30

402 4

1

8

1

1

1 0 0

-1

-1

-1


balance factors

27

Deletion Example (3) - continued


15

5 20

1

0

7

0

3 30

402 4

1

8

1

1

1 0 0

-1

-2

-1



1

Rotate 15

5

20

1

7

0

3

30

40

2 4

1

8

1

1

1 0 0

-1

Tree after RR imbalanceis corrected

0 0

0

Is this an AVL tree?

Continue backing up the tree

updating balance factors

Backup

28

Deletion Example (3) - continued



Rotate15

5

20

2

7

0

3

30

40

2 4

1

8

1

1

1 0 0

-1 0 0

0

Backup


5

7

0

3

2 4

1 8

0

1

1 0

0

15

20

30

400 0

0

0

-1

Final Tree

29

Search (Find)• Since AVL Tree is a BST, search algorithm

is the same as BST search and runs in guaranteed O(logn) time

30

Summary of AVL Trees• Arguments for using AVL trees:

1. Search/insertion/deletion is O(log N) since AVL trees are always balanced.

2. The height balancing adds no more than a constant factor to the speed of insertion/deletion.

• Arguments against using AVL trees:1. Requires extra space for balancing factor

(height)2. It may be OK to have a partially balanced tree

that would give performance similar to AVL trees without requiring the balancing factor• Splay trees

avl-trees: motivation

Documents