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CS350: Data Structures © James Moscola

College Catalog2009–2011

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James MoscolaDepartment of Engineering & Computer ScienceYork College of Pennsylvania

CS350: Data StructuresAVL Trees

CS350: Data Structures

Balanced Search Trees

• Binary search trees are not guaranteed to be balanced given random insertions and deletions

- Inserting a sorted lists of elements into a BST produces the worst case -- O(N)- Performance of an unbalanced tree can degrade as more elements are inserted

• Balanced search tree operations, such as insert, insure that the a tree always remains balanced

- An operation is not complete until it returns the tree to a balanced state

2

Balanced tree

Unbalanced tree


AVL Trees

• A type of balanced binary search tree

• Named for its discoverers -- Adelson, Velskii, and Landis

• Definition:

3

An AVL tree is a binary search tree with the additional balance property that, for any node in the tree, the height of the left and right subtrees can differ by at most 1. The height of an empty subtree is -1.


AVL Trees

4

12

4 10

8 16

14

2 60 0

1 0 0

12

3 The height of each node is listed in blue


-1


AVL Trees

5

12

4 10

8 16

14

2 60 0

1 0 0

12



-1

height differs by at most 1


AVL Trees

6

12

4 10

8 16

14

2 60 0

1 0 0

12



-1

height differs by at most 1


AVL Trees

7

12

4 10

8 16

14

2 6

1

0

1

-1

0

0

?

?

?

?

?

The binary tree after inserting node 1

What are the node heights?


AVL Trees

8

12

4 10

8 16

14

2 61 0

2 0 0

13

4

-1

10

The binary tree after inserting node 1

What are the node heights?


AVL Trees

9

12

4 10

8 16

14

2 61 0

2 0 0

13

4

-1

10

height differs bymore than 1

therefore, node 8 is invalid


AVL Trees

10

12

4 10

8 16

14

2 61 0

2 0 0

13

4

-1

10


therefore, node 12is unbalanced


therefore, node 8 is unbalanced


AVL Trees

• When a node in the tree no longer satisfies the invariant required to be an AVL tree the tree must be rebalanced around that node

• There are four ways in which an insertion into a tree may cause an imbalance to occur at some node X: (1) An insertion in the left subtree of the left child of X(2) An insertion in the right subtree of the left child of X(3) An insertion in the left subtree of the right child of X(4) An insertion in the right subtree of the right child of X

11

symmetric


Creating an Imbalance -- Case #1

12

A B

N1

C

N2

A

B

N1

C

N2

(1) An insertion in the left subtree of the left child of X

root of A subtree increases height

as does N1 and N2

Node N1 and root of C differ by more than 1 therefore N2

violates AVL rules

-1-10

-1-1 -10

1

21


Fixing the Imbalance -- Case #1

13

A

B

N1

C

Perform a single right rotation

N2

CB

N2

N1

A

N1 becomes new root node

N2 now right child of N1

B subtree now left child of N2

-1

-10

1

-1-1

0 0

12


Creating an Imbalance -- Case #4 (symmetric with 1)

14

CB

N1

A

N2

(4) An insertion in the right subtree of the right child of X

N1

C

B

N1

A

N2

root of C subtree increases height

as does N1 and N2

Node N1 and root of A differ by more than 1 therefore N2

violates AVL rules


Fixing the Imbalance -- Case #4 (symmetric with 1)

15

A B

N2

N1

C

N1 becomes new root node

N2 now left child of N1

B subtree now right child of N2

C

B

N1

A

N2

Perform a single left rotation


Creating an Imbalance -- Case #2

16

A

N1

C

N3

B

N1

C

N3

(2) An insertion in the right subtree of the left child of X

root of B subtree increases height

as does N1 and N3

B A

Node N1 and root of C differ by more than 1 therefore N3

violates AVL rules

-10

-1-1

-11

0-1

1 2



17

N1

C

N3

We know the node N2 exists because a node was just inserted into the B subtree

A

N2

B1

B2

N1

N3

A

C

B

-1

0-1

1

0-1

-11

-1 -1

2 2



18

N2

C

N3

B1

B2

N1

N3

A

CN2

B1

B2

Perform a left-right double rotation

A

N1

Step #1 - Perform a single left rotation between N1 and N2

tree is still unbalanced

N3

-10

-11

-1 -1

0-1

-11

-1 -1

22



19

N2

C

N3

B1

B2

A

N1

N1

C

N2

B1

B2A

Step #2 - Perform a single right rotation between N2 and N3

N3

-10

-11

-1 -1

-1 -1 -1 -1

0 0

12

Perform a left-right double rotation


Another look at the left-right double rotation without the intermediate steps


20

N1

C

N2

B1

B2A

N3N1

C

N3

A

N2

B1

B2


Here is a look at the right-left double rotation without the intermediate steps

Fixing the Imbalance -- Case #3 (symmetric with 2)

21

N3

C

N2

B1

B2A

N1N1

A

N3

C

N2

B2

B1


AVL Tree Insertion Example

22

12

4 10

8 16

14

2 60 0

1 0 0

12

3

-1


• Inserting node 1 - First, recursively search for the

location which to insert the node- Next, insert the node- Finally, unwind the recursion

update node heights along the way. Rebalance tree where necessary• To update node’s height,

check its children


23

12

4 10

8 16

14

2 6

1

0

1

-1

0

0

?

?

?

?

?



24

12

4 10

8 16

14

2 6

1

node 1 is a leaf node so its height is 0

0

1

-1

0

0

?

?

?

?

0



25

12

4 10

8 16

14

2 6

1

node 2 - add 1 to the height of its highest child

0

1

-1

0

0

1

?

?

?

0



26

12

4 10

8 16

14

2 6

1


0

1

-1

0

0

1

2

?

?

0



27

12

4 10

8 16

14

2 6

1

node 8 violates AVL properties, so rebalance


0

1

-1

0

0

1

2

?

?

0



28

12

4 10

8 16

14

2 6

1

Case #1: Must perform single right rotation between node 8

and node 4

0

1

-1

0

0

1

2

?

?

0



29

12

2 8

4 16

14

1

After rotation, update height of node 8 Node 4 is balanced, so continue

unwinding recursion

8

6 10

0

1

-1

0

1

0

1

2

?

0



30

12

2 8

4 16

14

1

8

6 10


0

1

-1

0

1

0

1

2

3

0


Other Operations: find / remove

• The find operation is the same as the unbalanced binary search tree

• The remove operation works similarly to the remove operation from the unbalanced binary search tree with a few modifications - When a node is removed, the heights of its ancestors may need to be

updated as the recursion is unwound -- fix imbalances as they are encountered just like with insertion

31


• Time complexity of AVL Tree operations

Analysis of AVL Tree Operations

32

worst case average

find O(log N) O(log N)

insert O(log N) O(log N)

remove O(log N) O(log N)

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