avl tree

AVL Tree

27th Mar 2007

AVL Trees Pencarian node pada unbalanced Binary Search

Trees adalah tidak efisien. Worst case: operations take O(n).

AVL (Adelson-Velskii & Landis) trees adalah BST yang diseimbangkan

Untuk setiap node pada tree, beda tinggi antara left subtree dan right subtree maximum 1 saja.

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X X

H

H-1H-2

AVL Trees

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12

8 16

4 10

2 6

14

Untuk setiap node pada tree, beda tinggi antara left subtree dan right subtree maximum 1 saja.

27th Mar 2007

AVL property violated here

AVL Trees

27th Mar 2007

10

5

3

20

2

1 3

10

5

3

20

1

43

5

Insertion for AVL TreeAfter insert 1

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12

8 16

4 10

2 6

14

1

Insertion for AVL Tree

Untuk memastikan AVL-tree dalam kondisi seimbang, setelah suatu node di-insert-kan, perlu dicek keseimbangan untuk tiap node

Jika setelah insertion ada node yang tidak seimbang, maka dilakukan operasi berikut :Single rotation

Double rotation

27th Mar 2007

Insertions Causing Imbalance

Kemungkinan insertion yang menyebabkan ketidakseimbangan Insertion pada P (outside) case1 Insertion pada Q (inside) case2

27th Mar 2007

RP

Q

k1

k2

Q

k2

P

k1

R

AVL TreeHP=HQ=HR

Kemungkinan insertion yang menyebabkan ketidakseimbangan Insertion pada Q (inside) case 3 Insertion pada R (outside) case 4

Single Rotation (case 1)

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A

k2

B

k1

CCBA

k1

k2

insertion pada A menyebabkan ketidakseimbangan

Single Rotation (case 4)

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C

k1

B

k2

A

A B C

k2

k1

insertion pada C menyebabkan ketidakseimbangan

Problem with Single Rotation Single rotation does not work for case 2 and 3

(inside case)

27th Mar 2007

Q

k2

P

k1

RR

P

Q

k1

k2

Insertion pada Q kemudian single rotation, tetap tidak seimbang

Double Rotation: Stepsama dengan dua kali single rotation

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C

k3

A

k1

D

B

k2

C

k3

A

k1

D

B

k2

Double Rotation: Step

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C

k3

A

k1

DB

k2

Double Rotation

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C

k3

A

k1

D

B

k2

C

k3

A

k1

DB

k2

Double Rotation

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B

k1

D

k3

A

C

k2

B

k1

D

k3

A C

k2

ExampleInsert 3 into the AVL tree

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3

11

8 20

4 16 27

8

8

11

4 20

3 16 27

ExampleInsert 5 into the AVL tree

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5

11

8 20

4 16 27 8

11

5 20

4 16 27

8

AVL Trees: Exercise

Insertion order:10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40,

5, 55

27th Mar 2007

Remove Operation in AVL Tree Removing a node from an AVL Tree is the

same as removing from a binary search tree. However, it may unbalance the tree.

Similar to insertion, starting from the removed node we check all the nodes in the path up to the root for the first unbalance node.

Use the appropriate single or double rotation to balance the tree.

May need to continue searching for unbalanced nodes all the way to the root.

27th Mar 2007

Deletion X in AVL Trees

Deletion:Case 1: if X is a leaf, delete XCase 2: if X has 1 child, use it to replace XCase 3: if X has 2 children, replace X with its

inorder predecessor (and recursively delete it)

Rebalancing

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Delete 55 (case 1)

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60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 55 (case 1)

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60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 50 (case 2)

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60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 50 (case 2)

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60

20 70

10 40 65 85

5 15 30 50 80 90

55

Delete 60 (case 3)

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60

20 70

10 40 65 85

5 15 30 50 80 90

55

prev

Delete 60 (case 3)

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55

20 70

10 40 65 85

5 15 30 50 80 90

Delete 55 (case 3)

27th Mar 2007

55

20 70

10 40 65 85

5 15 30 50 80 90

prev

Delete 55 (case 3)

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50

20 70

10 40 65 85

5 15 30 80 90

Delete 50 (case 3)

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50

20 70

10 40 65 85

5 15 30 80 90

prev

Delete 50 (case 3)

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40

20 70

10 30 65 85

5 15 80 90

Delete 40 (case 3)

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40

20 70

10 30 65 85

5 15 80 90

prev

Delete 40 : Rebalancing

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30

20 70

10 65 85

5 15 80 90

Case ?

Delete 40: after rebalancing

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30

7010

20 65 855

15 80 90

Single rotation is preferred!

Minimum Element in AVL Tree An AVL Tree of height H has at least FH+3-1 nodes,

where Fi is the i-th fibonacci number S0 = 1 S1 = 2 SH = SH-1 + SH-2 + 1

27th Mar 2007

SH-1

SH-2

H

H-1H-2

Summary

Find element, insert element, and remove element operations all have complexity O(log n) for worst case

Insert operation: top-down insertion and bottom up balancing

27th Mar 2007