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CSE 326: Data Structures Trees

Lecture 7: Wednesday, Jan 23, 2003

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Outline

• Finish discussion on random binary search trees (BST)

• AVL trees

• Reading assignment for this week:Weiss: 4.3, 4.4, 4.5, and 4.7

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The Average Depth of a BST

• Insert the elements 1 <2 < ... < n in some order, starting with the empty tree

• For each permutation, :– T = the BST after inserting (1), (2) , ... ,

(n)

• The Average Depth:))/n!height(T(H(n)

ππ

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The Average Depth of a BST

• The average depth of a BST is:

H(n) = (log n)

• For some , height(T) = O(log n)• For other , height(T) = O(n)• But the average is O(log n)

• Please read the proof in the book and/or slides !

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log versus n n

Why is average depth of BST's made from

random inputs different from the average

depth of all pos

log

Because there are more ways to build shallow

trees than d

sible

eep t

BST

ee

?

r

's

n n

s!

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Random Input vs. Random Trees

Inputs

1,2,3

3,2,1

1,3,2

3,1,2

2,1,3

2,3,1

TreesFor three items, the shallowest tree is twice as likely as any other – effect

grows as n increases. For n=4,

probability of getting a shallow

tree > 50%

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Average cost

• The average, amortized cost of n insert/find operations is O(log(n))

• But the average, amortized cost of n insert/find/delete operations can be as bad as (n)– Deletions make life harder (recall stretchy arrays)

• Need guaranteed cost O(log n) – next

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Beauty is Only (log n) Deep

• Binary Search Trees are fast if they’re shallowe.g.: complete

• Problems occur when one branch is much longer than the otherHow to capture the notion of a “sort of” complete

tree?

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Balance

balance = height(left subtree) - height(right subtree)• convention: height of a “null” subtree is -1• zero everywhere perfectly balanced• small everywhere balanced enough: (log n)

– Precisely: Maximum depth is 1.44 log n

t

56

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AVL Tree (Adelson-Velskii Landis)

4

121062

115

8

14137 9

• Binary search tree properties

• Balance of every node is -1 b 1

• Tree re-balances itself after every insert or delete

15What is the balance of each node in this tree?

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AVL Tree Data Structure

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92 12

5

10

20

17

0

0

100

1 2

3 10

3

data

height

children

300

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Not An AVL Tree

15

92 12

5

10

20

17

0

1

200

1 3

4 10

4

data

height

children

300

180

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Bad Case #1

Insert(small)

Insert(middle)

Insert(tall)

T

M

S

0

1

2

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Single Rotation

T

M

S

0

1

2

M

S T00

1

Basic operation used in AVL trees:

A right child could legally have its parent as its left child.

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General Case: Insert Unbalancesa

X

Y

b

Z

h h - 1

h + 1 h - 1

h + 2a

X

Y

b

Zh-1 h - 1

h h - 1

h + 1

a

XY

b

Z

h

h - 1

h

h - 1

h + 1

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Properties of General Insert + Single Rotation

• Restores balance to a lowest point in tree where imbalance occurs

• After rotation, height of the subtree (in the example, h+1) is the same as it was before the insert that imbalanced it

• Thus, no further rotations are needed anywhere in the tree!

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Bad Case #2

Insert(small)

Insert(tall)

Insert(middle)

M

T

S

0

1

2

Why won’t a single rotation (bringing T up to the top) fix this?

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Double Rotation

M

S T00

1

M

T

S

0

1

2

T

M

S

0

1

2

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General Double Rotation

• Initially: insert into X unbalances tree (root height goes to h+3)• “Zig zag” to pull up c – restores root height to h+2, left subtree height

to h

a

Z

b

W

c

Y

a

Z

b

W

c

Y

h+1

h

h

h

h + 3

h + 2

hh

h+1

h + 2

h+1

h

XX

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Another Double Rotation Case

• Initially: insert into Y unbalances tree (root height goes to h+2)• “Zig zag” to pull up c – restores root height to h+1, left subtree height

to h

a

Z

b

W

c

Y

a

Z

b

W

c

Y

h+1

h

h

h

h + 3

h + 2

hh

h+1

h + 2

h+1

h

XX

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Insert Algorithm

• Find spot for value

• Hang new node

• Search back up looking for imbalance

• If there is an imbalance:“outside”: Perform single rotation and exit

“inside”: Perform double rotation and exit

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AVL Insert AlgorithmNode insert(Comparable x, Node root){if ( root == NULL ) return new Node(x);if (x == root.key) return root;

if (x < root.key){root.left = insert( x, root.left );if (root unbalanced) { rotate... } }

else { // x > root.keyroot.right = insert( x,

root.right ); if (root unbalanced) { rotate... } }

root.height = max(root.left.height, root.right.height)+1;

return root;}

Node insert(Comparable x, Node root){if ( root == NULL ) return new Node(x);if (x == root.key) return root;

if (x < root.key){root.left = insert( x, root.left );if (root unbalanced) { rotate... } }

else { // x > root.keyroot.right = insert( x,

root.right ); if (root unbalanced) { rotate... } }

root.height = max(root.left.height, root.right.height)+1;

return root;}

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Deletion (Really Easy Case)

2092

155

10

30173

121

0

100

2 2

3

00

Delete(17)

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Deletion (Pretty Easy Case)

2092

155

10

30173

121

0

100

2 2

3

00

Delete(15)

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Deletion (Pretty Easy Case cont.)

2092

175

10

303

121 100

2 2

3

00

Delete(15)

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Deletion (Hard Case #1)

2092

175

10

303

121 100

2 2

3

00

Delete(12)

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Single Rotation on Deletion

2092

175

10

303

1 10

2 2

3

00

3092

205

10

17

3

1 00

2 1

3

0

0

What is different about deletion than insertion?

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Deletion (Hard Case)

Delete(9)

2092

175

10

303

121 220

2 3

4

0

33

15

130 0

1

0

20

30

12

33

15

13

1

0 0

110

180

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Double Rotation on Deletion

2

3

0

202

175

10

30

121 22

2 3

4

33

15

13

1

0 0

111

0180

2052

173

10

30

120 220

1 3

4

33

15

13

1

0 0

111

01800

Not finished!

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Deletion with Propagation

We get to choose whether to single or double rotate!

2052

173

10

30

120 220

1 3

4

33

15

13

1

0 0

111

0180

What different about this case?

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Propagated Single Rotation

0

30

20

17

33

12

15

13

1

0

52

3

10

4

3 2

1 2 1

0 0 011

0

2052

173

10

30

120 220

1 3

4

33

15

13

1

0

111

0180

180

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Propagated Double Rotation

0

17

12

11

52

3

10

4

2 3

1 0

0 0

2052

173

10

30

120 220

1 3

4

33

15

13

1

0

111

0180

151

0

20

30

33

1180

130

2

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AVL Deletion Algorithm• Recursive1. If at node, delete it

2. Otherwise recurse to find it in

3. Correct heights

a. If imbalance #1,

single rotate

b. If imbalance #2 (or don’t

care),

double rotate

• Iterative1. Search downward for node, stacking parent nodes2. Delete node3. Unwind stack, correcting heights a. If imbalance #1, single rotate b. If imbalance #2

(or don’t care) double rotate

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Pro:

• All operations guaranteed O(log N) • The height balancing adds no more than a

constant factor to the speed of insertion

Con:

• Space consumed by height field in each node• Slower than ordinary BST on random data

Can we guarantee O(log N) performance with less overhead? Splay trees next time

Pros and Cons of AVL Trees