foundation of computing systems

Foundation of Computing Systems

Lecture 8

Trees: Part V

12.08.09 IT 60101: Lecture #8 2

Height Balanced Binary Trees

• Average search time in a binary search tree

• A binary search tree should be with a minimum value of average search time

n

n

ii

1

i = Number of comparisons for the i-th element

n = Total number of elements in the binary search tree

12.08.09 IT 60101: Lecture #8 3

Some Sample Binary Search Trees

A skew ed b ina ry tree ob ta ined from lexicog raph ica l o rde r o f da ta

apraug

decfeb

janju l

junm ar

m aynov

oct

sep

= 6.50

12.08.09 IT 60101: Lecture #8 4


= 4.00

jun

ju l

jan

feb

dec

aug

apr

m ar

m ay

nov

oct

sep

A b ina ry search tree ( ha lf skew ed ve rs ion )

jan

feb m ar

apr m ayjun

ju laug

dec

sep

oct

nov

A b ina ry search tree (ob ta ined by inse rting the

da ta in to the o rde r o f m on ths ).

= 3.50

12.08.09 IT 60101: Lecture #8 5


= 3.08 = 3.08

jan

feb

m ar

apr

m ayjun

ju laug

dec

sep

oct

nov

B ina ry sea rch tree in the fo rm o f a com p le teb ina ry tree

jan

feb

m ar

apr

m ay

jun

ju l

aug

dec

sep

oct

nov

A b ina ry sea rch tree ob ta ined from a g iven

random o rdering o f da ta .

12.08.09 IT 60101: Lecture #8 6


= 3.16

jan

feb

m ar

apr

m ayjun

ju laug

dec

sep

oct

nov

A b ina ry sea rch tree ob ta ined by a spec ia l techn ique

12.08.09 IT 60101: Lecture #8 7


• How to find a binary search tree with a minimum value of average search time ?

• Solution• Height balanced binary search tree

• Also called AVL tree

• Concept of balance factor of a node

bf = Height of the left sub-tree (hL) – Height of the right sub-tree (hR)

12.08.09 IT 60101: Lecture #8 8


• Definition

A binary search tree is said to be height balanced binary search tree

if all its nodes have a balance factor of 1, 0 or –1.

That is, for all nodes

|bf| = |hL – hR| ≤ 1

12.08.09 IT 60101: Lecture #8 9

Height Balanced Binary Trees: Example

6

2

1

3

4 7

8

1

0

1

10 0

(a) Height balanced

6

2

1

3

4

5

8

2

0

0

00

0

(b) Height unbalanced

12.08.09 IT 60101: Lecture #8 10

Managing Unbalanced BST

• Steps

• Insert node into a binary search tree

• Compute the balance factors

• Decide the pivot node

• Balance the unbalance tree

12.08.09 IT 60101: Lecture #8 11

Unbalanced to Balanced BST

• AVL Rotations• G.M. Adelson-Velskii and E.M. Lendis (1962)

• Case 1: Left-to-Left rotation• pivotleft-sub treeleft child

• Case 2: Right-to-Right rotation• pivotright-sub treeright child

• Case 3: Left-to-Right rotation• pivotleft-sub treeright child

• Case 4: Right-to-Left rotation• pivotright-sub treeleft child

12.08.09 IT 60101: Lecture #8 12

Left-to-Left Rotation

P

A

A A

P

L R

R

P

A

A

A P

L

R R

A fter ro ta tionA fte r insertion inthe le ft sub-tree o fthe le ft ch ild o f p ivo t node (P )

12.08.09 IT 60101: Lecture #8 13


6

3

4 7

81

1

0

0 0

5

0

0

0

9

0

11

106

3

4 7

81

1

0

0 0

5

0

0

0

9

0

11

10

2

2

1

1

2

0

(b) A fter the insertion o f 2 in to the tree(a) A he ight ba lanced b inary tree as |b f| 1

12.08.09 IT 60101: Lecture #8 14


6

3

4 7

8

50

9

0

11

10

2

P

A

A

A

P

L

R

R

(c) A V L-ro ta tion as per the LE FT-TO -LE FT(C ase 1) insertion

63

4

7

8

5

9 11

10

2

PRA

R

AL

(d) A fter A V L-ro tation

12.08.09 IT 60101: Lecture #8 15


3

4

6

81

1

0

0

0

5

0

0

0

9

0

11

10

2 70

(e) H e ight ba lanced tree w ith new ly inserted node 2 and a fter A VL-ro ta tion

12.08.09 IT 60101: Lecture #8 16

Right-to-Right Rotation

P

B

BB

P

RL

L

P

B

P B

B

L L

R

A fter ro ta tionA fte r inse rtion in the righ t sub -treeo f the righ t ch ild o f the p ivo t node (P )

12.08.09 IT 60101: Lecture #8 17

Right-to-Right Rotation

6

13

4

7

8

-1

1

0

00

5

0

0

9

0

11

10

12

-2

- -2

-1

P

B

P

B

B L

L

R

0

(a ) 12 is inserted wh ich m akes the tree unbalanced

613

4

7

8 0

0

5

0

0

9

0

11

10 12

B

P

B

BL R

0

-1

0

0

PL

P

L

(b) A fte r A V L ro ta tion

12.08.09 IT 60101: Lecture #8 18

Left-to-Right Rotation

• Also called double rotation• Rotation 1 (Right-to-Left)

– Left sub-tree (BL) of the right child (B) of the left child of the pivot node (P) becomes the right sub-tree of the left child (A).

– Left child (A) of the pivot node (P) becomes the left child of B.

• Rotation 2 (Left-to-Right)– Right sub-tree (BR) of the right child (B) of the left child

(A) of the PIVOT node (P) becomes the left sub-tree of P.

– P becomes the right child of B.

12.08.09 IT 60101: Lecture #8 19


P

A

A

B

P

L

R

R

A fte r inse rtion in the righ t sub-treeo f the le ft ch ild o f the p ivo t node P After ro ta tion

B1

2

BL

PA

A B PL R R

B

BL

12.08.09 IT 60101: Lecture #8 20


3

2

81

0

1 00

0

0

9

0

11

10

(a) A he ight ba lanced tree

3

7

81

1

0

0

5

0

0

90

11

10

2

1

(b) Insertion o f 7 m ade the tree as unba lanced

1

5

0

4 6

00-

2

1

-1

00

4 6

-1

0

Balance factors are recom puted

12.08.09 IT 60101: Lecture #8 21


3

7

8

0

5

0

0

9 11

10

2

4 6

0A

L

PR

BR

P

B

A

(c) N ode 8 becom es the P IV O T node

3

2

8

9 11

10

1

5

4 6

AL B

L

BL

BR P

R

(d ) A fte r A V L-ro ta tion

11

12.08.09 IT 60101: Lecture #8 22

Right-to-Left Rotation

• It is also another double rotation• Rotation 1(Left-to-Right)

– Right sub-tree (BR) of the left child (B) of the right child (A) of the pivot node (P) becomes the left sub-tree of A.

– Right child (A) of the pivot node (P) becomes the right child of B.

• Rotation 2 (Right-to-Left)– Left sub-tree (BL) of the right child (B) of the right child

(A) of the pivot rode (P) becomes the right sub-tree of P.– P becomes the left child of B.

12.08.09 IT 60101: Lecture #8 23


A fte r ro ta tion

P A

ABPL RR

B

BL

P

A

A

B

PL

R

R

B 1

2

BL

A fte r inse rtion in the le ft sub -treethe righ t ch ild o f the p ivo t node (P )

12.08.09 IT 60101: Lecture #8 24


7

80

5

0

09

124 6

0

A

L

P

R

B R

P

2

A

B

L 1311

10

0

00

-1-2

-1-2

01

(a ) W hen 10 is inse rted in to the tree

7

8

5

9 124 6

AL R

BR

B

13

11

10

PL

(b ) A fte r A V L ro ta tions

1

12.08.09 IT 60101: Lecture #8 25


(c ) F ina l he igh t ba lanced tree a fte r A V L ro ta tion

7

8

5

9 124 6

13

11

10

-1

0

00

0

0

0

0

-1

0

12.08.09 IT 60101: Lecture #8 26

Implementation

For tricky ways to manage the pointers see the bookClassic Data Structures

Chapter 7PHI, 2nd Edn., 17th Reprint

12.08.09 IT 60101: Lecture #8 27

Property of a HBT

• Maximum height possible in an AVL tree with n number

of nodes is given by

For the proof of this property see the bookClassic Data Structures

Chapter 7PHI, 2nd Edn., 17th Reprint

nh 2max log44.1

foundation of computing systems

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