avl trees

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AVL Trees 1 © 2004 Goodrich, Tamassia AVL Trees 6 3 8 4 v z

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Page 1: Avl trees

AVL Trees 1© 2004 Goodrich, Tamassia

AVL Trees6

3 8

4

v

z

Page 2: Avl trees

AVL Trees 2© 2004 Goodrich, Tamassia

AVL Tree Definition

AVL trees are balanced.An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1.

88

44

17 78

32 50

48 62

2

4

1

1

2

3

1

1

An example of an AVL tree where the heights are shown next to the nodes:

Page 3: Avl trees

AVL Trees 3© 2004 Goodrich, Tamassia

Height of an AVL TreeFact: The height of an AVL tree storing n keys is O(log n).Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h.We easily see that n(1) = 1 and n(2) = 2For n > 2, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and another of height n-2.That is, n(h) = 1 + n(h-1) + n(h-2)Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). Son(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction),n(h) > 2in(h-2i)

Solving the base case we get: n(h) > 2 h/2-1

Taking logarithms: h < 2log n(h) +2Thus the height of an AVL tree is O(log n)

3

4 n(1)

n(2)

Page 4: Avl trees

AVL Trees 4© 2004 Goodrich, Tamassia

Insertion in an AVL TreeInsertion is as in a binary search treeAlways done by expanding an external node.Example: 44

17 78

32 50 88

48 62

54w

b=x

a=y

c=z

44

17 78

32 50 88

48 62

before insertion after insertion

Page 5: Avl trees

AVL Trees 5© 2004 Goodrich, Tamassia

Trinode Restructuringlet (a,b,c) be an inorder listing of x, y, zperform the rotations needed to make b the topmost node of the three

b=y

a=z

c=x

T0

T1

T2 T3

b=y

a=z c=x

T0 T1 T2 T3

c=y

b=x

a=z

T0

T1 T2

T3b=x

c=ya=z

T0 T1 T2 T3

case 1: single rotation(a left rotation about a)

case 2: double rotation(a right rotation about c, then a left rotation about a)

(other two cases are symmetrical)

Page 6: Avl trees

AVL Trees 6© 2004 Goodrich, Tamassia

Insertion Example, continued

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32 50

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2

5

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1

T0T2

T3

x

y

z

2

3

4

5

67

1

88

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17

7832 50

48

622

4

1

1

2 2

3

154

1

T0 T1

T2

T3

x

y z

unbalanced...

...balanced

1

2

3

4

5

6

7

T1

Page 7: Avl trees

AVL Trees 7© 2004 Goodrich, Tamassia

Restructuring (as Single Rotations)

Single Rotations:

T0T1

T2

T3

c = xb = y

a = z

T0 T1 T2

T3

c = xb = y

a = zsingle rotation

T3T2

T1

T0

a = xb = y

c = z

T0T1T2

T3

a = xb = y

c = zsingle rotation

Page 8: Avl trees

AVL Trees 8© 2004 Goodrich, Tamassia

Restructuring (as Double Rotations)

double rotations:

double rotationa = z

b = xc = y

T0T2

T1

T3 T0

T2T3T1

a = zb = x

c = y

double rotationc = z

b = xa = y

T0T2

T1

T3 T0

T2T3 T1

c = zb = x

a = y

Page 9: Avl trees

AVL Trees 9© 2004 Goodrich, Tamassia

Removal in an AVL TreeRemoval begins as in a binary search tree, which means the node removed will become an empty external node. Its parent, w, may cause an imbalance.Example:

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7832 50

8848

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54

44

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7850

8848

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54

before deletion of 32 after deletion

Page 10: Avl trees

AVL Trees 10© 2004 Goodrich, Tamassia

Rebalancing after a Removal

Let z be the first unbalanced node encountered while travelling up the tree from w. Also, let y be the child of z with the larger height, and let x be the child of y with the larger height.We perform restructure(x) to restore balance at z.As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached44

17

7850

8848

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54

w

c=x

b=y

a=z

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50 88

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Page 11: Avl trees

AVL Trees 11© 2004 Goodrich, Tamassia

Running Times for AVL Trees

a single restructure is O(1) using a linked-structure binary tree

find is O(log n) height of tree is O(log n), no restructures needed

insert is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log

n)

remove is O(log n) initial find is O(log n) Restructuring up the tree, maintaining heights is O(log

n)