analysis algorithm 8

7/27/2019 Analysis Algorithm 8

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AVL Trees

1AVL TREE


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Overview

Binary treesBinary search trees

Find

Insert

Delete

AVL trees

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Hierarchical Data Structures

Tree

Single parent, multiple children

Binary tree

Tree with 02 children per node

Tree Binary Tree3


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Trees

TerminologyRoot no predecessorLeaf no successorInterior non-leaf

Height distance from root to leaf

Root node

Leaf nodes

Interior nodes Height

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Types of Binary Trees

Degenerate

only one childBalanced mostly two children

Complete always two children

Degenerate

binary tree

Balanced

binary tree

Complete

binary tree5


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Binary Trees Properties

DegenerateHeight = O(n) for n

nodes

Similar to linear list

BalancedHeight = O( log(n) )

for n nodes

Useful for searches

Degenerate

binary tree

Balanced

binary tree6


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Binary Search Trees

Key propertyValue at node

Smaller values in left subtree

Larger values in right subtree

Example

X >Y

X < Z

Y

X

Z

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Binary Search Trees

Examples

Binary

search trees

Non-binary

search tree

5

10

30

2 25 45

5

10

45

2 25 30

5

10

30

2

25

45

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Example Binary Searches

Find ( 2 )

5

10

30

2 25 45

5

10

30

2

25

45

10 > 2, left

5 > 2, left2 = 2, found

5 > 2, left

2 = 2, found

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Binary Search Properties

Time of search

Proportional to height of tree

Balanced binary tree

O( log(n) ) time

Degenerate tree

O( n ) time

Like searching linked list / unsorted array

RequiresAbility to compare key values

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Binary Search Tree Insertion

Algorithm

Perform search for value X

Search will end at node Y (if X not in tree)

If X < Y, insert new leaf X as new left subtree for Y

If X > Y, insert new leaf X as new right subtree for Y

Observations

O( log(n) ) operation for balanced tree

Insertions may unbalance tree

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Example Insertion

Insert ( 20 )

5

10

30

2 25 45

10 < 20, right

30 > 20, left

25 > 20, left

Insert 20 on left

20

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Binary Search Tree Deletion

Algorithm

Perform search for value X

If X is a leaf, delete X

Else // must delete internal node

a) Replace with largest value Y on left subtreeOR smallest value Z on right subtree

b) Delete replacement value (Y or Z) fromsubtree

ObservationO( log(n) ) operation for balanced tree

Deletions may unbalance tree

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Example Deletion (Leaf)

Delete ( 25 )

5

10

30

2 25 45

10 < 25, right

30 > 25, left

25 = 25, delete

5

10

30

2 45

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Example Deletion (Internal Node)

Delete ( 10 )

5

10

30

2 25 45

5

5

30

2 25 45

2

5

30

2 25 45

Replacing 10

with largest

value in left

subtree

Replacing 5

with largest

value in left

subtree

Deleting leaf

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Example Deletion (Internal Node)

Delete ( 10 )

5

10

30

2 25 45

5

25

30

2 25 45

5

25

30

2 45

Replacing 10

with smallest

value in right

subtree

Deleting leaf Resulting tree

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17/60AVL TREE

17

AVL - Good but not Perfect Balance

AVL trees are height-balanced binarysearch trees

Balance factorof a nodeheight(left subtree) - height(right subtree)

An AVL tree has balance factor calculated atevery node

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

Store current heights in each node


18/60AVL TREE

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Node Heights

1

00

2

0

6

4 9

81 5

1

height of node = h

balance factor = hleft-hrightempty height = -1

0

0

height=2 BF=1-0=1

0

6

4 9

1 5

1

Tree A

(AVL)Tree B

(AVL)


19/60AVL TREE

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Node Heights after Insert 7

2

10

3

0

6

4 9

81 5

1

height of node = h

balance factor = hleft-hrightempty height = -1

1

0

2

0

6

4 9

1 5

1

07

07

balance factor

1-(-1) = 2

-1

Tree A

(AVL)Tree B (not

AVL)


20/60AVL TREE

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Insert and Rotation in AVL Trees

Insert operation may cause balance factor tobecome 2 or2 for some node

only nodes on the path from insertion point toroot node have possibly changed in height

So after the Insert, go back up to the root node bynode, updating heights

If a new balance factor (the difference hleft-hright) is

2 or

2, adjust tree by rotat ion around the node


21/60AVL TREE

21

Single Rotation in an AVL Tree

2

10

2

0

6

4 9

81 5

1

0

7

0

1

0

2

0

6

4

9

8

1 5

1

07


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AVL TREE

22

Let the node that needs rebalancing be .

There are 4 cases:

Outside Cases (require single rotation) :

1. Insertion into left subtree of left child of.2. Insertion into right subtree of right child of.

Inside Cases (require double rotation) :

3. Insertion into right subtree of left child of.4. Insertion into left subtree of right child of.

The rebalancing is performed through four

separate rotation algorithms.

Insertions in AVL Trees


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AVL TREE

23

j

k

X Y

Z

Consider a valid

AVL subtree

AVL Insertion: Outside Case

h

hh


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AVL TREE

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j

k

XY

Z

Inserting into X

destroys the AVL

property at node j


h

h+1 h


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AVL TREE

25

j

k

XY

Z

Do a right rotation


h

h+1 h


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AVL TREE

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j

k

XY

Z

Do a right rotation

Single right rotation

h

h+1 h


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AVL TREE

27

j

k

X Y Z

Right rotation done!

(Left rotation is mirror

symmetric)

Outside Case Completed

AVL property has been restored!

h

h+1

h


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AVL TREE

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j

k

X Y

Z

AVL Insertion: Inside Case

Consider a valid

AVL subtree

h

hh


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AVL TREE

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Inserting into Y

destroys the

AVL property

at node j

j

k

X Y

Z


Does right rotation

restore balance?

h

h+1h


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AVL TREE30

j

k

X

Y Z

Right rotation

does not restore

balance now k is

out of balance


hh+1

h


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AVL TREE31

Consider the structure

of subtree Y j

k

X Y

Z


h

h+1h


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AVL TREE32

j

k

XV

Z

W

i

Y = node i and

subtrees V and W


h

h+1h

h or h-1


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AVL TREE33

j

k

XV

Z

W

i


We will do a left-right

double rotation . . .


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AVL TREE34

j

k

X V

Z

W

i

Double rotation : first rotation

left rotation complete


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AVL TREE35

j

k

X V

Z

W

i

Double rotation : second

rotationNow do a right rotation


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AVL TREE36

jk

X V ZW

i

Double rotation : second

rotationright rotation complete

Balance has been

restored

hh h or h-1


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AVL TREE37

Example of Insertions in an AVL Tree

1

0

220

10 30

25

0

35

0

Insert 5, 40


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AVL TREE39

Single rotation (outside case)

2

0

320

10 30

25

1

35

2

50

20

10 30

25

1

405

40

0

0

0

1

2

3

45

Imbalance

3545

0 0

1

Now Insert

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AVL TREE40

Double rotation (inside case)

3

0

320

10 30

25

1

40

2

50

20

10 35

30

1

405

45

0 1

2

3

Imbalan

ce

45

0

1

Insertion of

34

35

34

0

0

125 34

0


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5

3

1 4

Insert 3.5

AVL Tree

8

3.5

5

3

1 4

8

4

5

1

3

3.5 After Rotation

x

y

A z

B

C

8

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An Extended Example

Insert 3,2,1,4,5,6,7, 16,15,14

3

Fig 1

3

2

Fig 2

3

2

1

Fig 3

2

1 3Fig 4

2

1 3

4

Fig 5

2

1 3

4

5

Fig 6

Single rotation

Single rotation

42

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2

1 4

53

Fig 7 6

2

1 4

53

Fig 8

4

2 5

61 3

Fig 9

4

2 5

61 3

7Fig 10

4

2 6

71 3

5 Fig 11

Single rotation

Single rotation

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4

2 6

71 3

5 16

Fig 12

4

2 6

71 3

5 16

15Fig 13

4

2 6

151 3 5

167Fig 14

Double rotation

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5

4

2 7

151 3

6

1614

Fig 16

4

2 6

15

13

516

7

14

Fig 15

Double rotation

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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 tothe 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.

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Deletion X in AVL Trees

Deletion:

Case 1: if X is a leaf, delete X

Case 2: if X has 1 child, use it to replace X

Case 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)

60

20 70

10 40 65 85

5 15 30 50 80 90

55

48

D l ( 1)


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Delete 55 (case 1)

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20 70

10 40 65 85

5 15 30 50 80 90

55

49

D l t 50 ( 2)


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Delete 50 (case 2)

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20 70

10 40 65 85

5 15 30 50 80 90

55

50

D l t 50 ( 2)


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Delete 50 (case 2)

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20 70

10 40 65 85

5 15 3050 80 90

55

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D l t 60 ( 3)


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Delete 60 (case 3)

60

20 70

10 40 65 85

5 15 30 50 80 90

55

prev

52

D l t 60 ( 3)


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Delete 60 (case 3)

55

20 70

10 40 65 85

5 15 30 50 80 90

53

D l t 55 ( 3)


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Delete 55 (case 3)

55

20 70

10 40 65 85

5 15 30 50 80 90

prev

54

D l t 55 ( 3)


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Delete 55 (case 3)

50

20 70

10 40 65 85

5 15 30 80 90

55

D l t 50 ( 3)


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Delete 50 (case 3)

50

20 70

10 40 65 85

5 15 30 80 90

prev

56

D l t 50 ( 3)


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Delete 50 (case 3)

40

20 70

10 30 65 85

5 15 80 90

57

D l t 40 ( 3)


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Delete 40 (case 3)

40

20 70

10 30 65 85

5 15 80 90

prev

58

Delete 40 Rebalancing


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Delete 40 : Rebalancing

30

20 70

10 65 85

5 15 80 90

Case ?

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analysis algorithm 8

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