prepared by:mitali sonar (lect. computer dep.)

PREPARED BY:Mitali sonar

(lect. Computer Dep.)

ASSIGNMENT-4 LINEAR DATA STRUCTURE QUEUE

• Explain Queue fundamentals with Queue Insertion &

deletion algorithms

• Representation of Queue using Array and link list.

• Describe Circular Queue Insertion & Deletion

algorithm with an example.

• Differentiate: Simple Queue Vs Circular Queue.

• Explain DeQueue and its types in detail

• Explain Priority Queue. Explain the use of it in

computer field.

• Differentiate: Simple Queue Vs Priority Queue

• Write Advantage & disadvantage of circular queue

over simple queue.

• Queue application – SIMULATION

• What is the difference between LIFO and FIFO?

SUBMIT ON

13/09/2010

A nonlinear data structure is one in which its

element do not form a sequence

Many time we observe hierarchical relationship

between various data elements. This hierarchical

relationship between data elements can be

easily represented using tree.

A tree consists multiple node with each node

containing zero ,one or more pointers to other

nodes.A

B C

D E F G

A tree is a nonlinear data structure representing

more elements are called NODE.

Each node of a tree stores data value & has zero

or more pointers pointing to other nodes are

known as CHILD NODE.

The node at top of tree is known as root node of

the tree.

Node at lowest level are known as LEAF NODE.

A

B C

D E F G

ROOT NODE

INTERMEDIATE

NODE

LEAF NODE

The PARENT NODE is immediate predecessor of a node Each child node has one parent node.

Any node having child node as well as parent node is known as INTERNAL NODE.

DEGREE:-The maximum number of children that is possible for a node is known as degree of a node.

SIBBLING the node which have same parent are called sibbling.D,E,F ARE sibbling

LEVEL each node belongs to particular level number.

A

B C

D E F G

Level 0

Level 1

Level 2

DEPTH –it’s a highest level no of any leaf node.in

fig node D,E,F,G has level 2 so its depth is 2. The

depth of a node n is the length of the path from

the root to the node.

The height of a tree is the length of the path from

the root to the deepest node in the tree. A

(rooted) tree with only one node (the root) has a

height of zero

ANCESTOR & DESCENDANT :-A node N1 is said to

ancestor of node N2 if N1 is parent node of N2 Or

parent of parent node of N2

Node N2 is said to be descendant of Node N1.The

node N2 is said left descendant of N1 if it is

belongs to left subtree and N2 is said right

descendant of N1 if it is belongs to right subtree

BINARY TREE:-A binary tree consist of set of nodes

that is either empty or has following properties

1. One of node is designated as root node

2. The remaining nodes are partitioned into two

disjoint subset called left subtree & right subtree

respectively each of which is a binary tree.

3. Each node in binary tree can have atmost 2

children only

Left

subtree

Right

subtree

FULL BINARY TREE:-A binary tree is full binary

tree (sometimes proper/perfect binary tree ) if

and only if

1. each non leaf node has exactly two child node

2. all leaf node are at same level.

STRICTLY BINARY TREE:-A binary tree is

strictly binary tree if and only if

Each node has exactly two child nodes or no

nodes

complete binary tree

A binary tree is a complete binary tree if and only if

Level 0 t o h-1 represent a full binary tree of height

h-1

One or more nodes in level h-1 may have 0, or 1

child nodes

EXTENDED BINARY TREE(2-TREE)

A binary tree is said to be extended binary

tree if all of its nodes has either zero or two

degree .

The nodes having zero degree known as

external node represented as squares & nodes

with degree two known as internal node

represented as circles.

Properties of binary trees

The number of nodes n in a perfect binary tree can be found using this formula: n = 2h + 1 − 1where h is the height of the tree.

The number of nodes n in a complete binary tree is minimum: n = 2h and maximum: n = 2h + 1 − 1where h is the height of the tree.

The number of nodes n in a perfect binary tree can also be found using this formula: n = 2L − 1where L is the number of leaf nodes in the tree.

The number of leaf nodes n in a perfect binary tree can be found using this formula: n = 2h where h is the height of the tree.

The number of NULL links in a Complete Binary Tree of n-node is (n+1).

The number of leaf node in a Complete Binary Tree of n-node is UpperBound(n / 2).

ARRAY REPRESENTATION

Binary tree is represented sequentially in memory

by using single one dimensional array. All the

node of a tree are assigned a sequence number.

Suppose we number the nodes from left to right,

beginning at the top and ending at the bottom.

Then we can store the various data items in the

corresponding elements of an array.

0

12

3 4

LINKED REPRESENTATION

The linked representation of binary tree is

implemented using a linked list having INFO part

& two Pointer

Info part contains data values & two pointers

left & right are used to point to left & right

subtree of a node respectively

0

12

3 4

42

40 35

10 31

NUll NUll NUll NUll

NUll NUll

Traversal is the process of visiting every node

once

Preoder, Inorder, Postorder

Preorder Traversal:

1. Visit the root

2. Traverse left subtree

3. Traverse right subtree

Inorder Traversal:


2. Visit the root


Postorder Traversal:



3. Visit the root

CONVERSE PREORDER – if the word left and

right are interchanged in preceding definition

we obtain converse preorderA

F

GE

DB

C

A D G E F B C

Process root node

Right node

Left Node

PREORDER(T)-

Given a binary tree whose root node address is

given by a pointer T.

S & TOP denotes stack & top index

P denotes current node in tree.

1.[Intialize]

if T= NULL

then write (‘EMPTY TREE’)

return

Else TOP 0

call PUSH(S,TOP,T)

2.[Process each stacked branch address]

Repeat step -3 while TOP>0

3.[get stored address & branch left]

P POP(S,TOP)

Repeat while P != NULL

write(DATA(P))

if RPTR(P)!=NULL

then call PUSH(S,TOP,RPTR(P))

PLPTR(P)

4.[Finished]

return

RPREORDER(T)

1.[Process the root node]

If T!= NULL

Then write(DATA(T))

Else write(‘EMPTY TREE’)

return

2.[Process the left subtree]

If LPTR(T)!= NULL

Then call RPREORDER(LPTR(T))

3.[Process the right subtree]

If RPTR(T)!=NULL

then call RPREORDER(RPTR(T))

4.[finished]

return

INORDER

Left node

Process root node

Right Node

a

b c

b a c

a

b c

d ef

g h i j

g d h b e i a f j c

INORDER

Inorder By Projection

a

b c

d ef

g h i j

g d h b e i a f j c

INORDER

A

F

GE

DB

C

C B A E F G D A

CONVERSE INORDER

Right node

Process root node

Left Node

Converse inorder

G D F E A B C

RINORDER(T) Given a binary tree whose root

node address is given by a pointer T.

S & TOP denotes stack & top index

P denotes current node in tree.

1.[Check for empty tree]

If T = NULL

Then write(‘EMPTY TREE’)

return


If LPTR(T)!= NULL

Then call RINORDER(LPTR(T))


write (DATA (T))


If RPTR(T)!=NULL

then call RINORDER(RPTR(T))

5.[finished]

return

POSTORDER

Left node

Right Node

Process root nodea

b c

d ef

g h i j

g h d i e b j f c a

POSTORDER

A

F

GE

DB

C

C B F E G D A

CONVERSE POSTORDER

Right node

Left Node

Process root node

Converse post order

G F E D C B A

POSTORDER(T)

1.[Intialize]

If T = NULL


return

Else P T

TOP 0

2.[Traverse in post order]

Repeat thru step 5 while true

3.[Traverse left]

Repeat while P != NULL

call PUSH(S,TOP,P)

PLPTR(P)

4.[Process a node whose left & right subtree

have been traversed]

Repeat while S[TOP] <0

P POP(S,TOP)

Write(DATA(P))

If TOP = 0

then return

5.[Branch right & then mark node from

which we branched]

P RPTR (S[TOP])

S[TOP] - S[TOP]

RPOSTORDER(T)

1.[Check for empty tree]

If T = NULL


return


If LPTR(T)!= NULL

Then call RPOSTORDER(LPTR(T))


If RPTR(T)!=NULL

then call RPOSTORDER(RPTR(T))


write (DATA (T))

5.[finished]

return

Formation of binary tree from its traversal

If preorder traversal is given then 1st node is

root node

If postorder is given then last node is root

INORDER:- D B H E A I F J C G

PREORDER : A B D E H C F I J G

A is root node

In inorder all nodes which are on leftside of A

belongs to left subtree & those which are on

right side of A belongs to right subtree.

Inorder: n1 n2 n3 n4 n5 n6 n7 n8 n9

Postorder: n1 n3 n5 n4 n2 n8 n7 n9 n6

n6

IN:- n1 n2 n3 n4 n5

Post:- n1 n3 n5 n4 n2

IN:- n7 n8 n9

Post:- n8 n7 n9

Obtain binary tree from given general ordered

tree.

fb

a

c d g j k

e h i

A

B C D

K H I J E F

G

Conversion of general tree to binary tree

BINARY SEARCH TREE a binary search tree

(BST) has the following

properties:

The left subtree of a node

contains only nodes with

keys less than the node's

key.

The right subtree of a

node contains only nodes

with keys greater than the

node's key.

Both the left and right

subtrees must also be

binary search trees

CREATION OF BINARY SEARCH TREE

40,15,65,35,5,45,75,95,85,05,30

MARCH,MAY,NOV,AUG,APR,JAN,DEC,JUL,FEB,

JUN,OCT & SEP.

Searching BST If we are searching for 15, then node is

found

If we are searching for a key < 15, then we

should search in the left subtree.

If we are searching for a key > 15, then we

should search in the right subtree.

Example Insertion

Insert ( 20 )

5

10

30

2 25 45

20 >10, right

20 < 30, left

20 < 25, left

Insert 20 on left

20

10

30

Example Deletion (Leaf)

Three cases:

(1) the node is a leaf

– Delete it immediately

– Delete ( 25 )

5

10

30

2 25 45

25 > 10, right

25 < 30, left

25 = 25, delete

5

10

30

2 45

Example Deletion (Internal Node)

(2) the node has one childAdjust a pointer from the parent to bypass that node.

• Delete ( 5 )

5

10

30

2 25 45

2

10

30

2 25 45

delete(3) the node has 2 children

replace the key of that node with the minimum element at the right subtree

delete the minimum element

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

THREADED BINARY TREE:-

In a binary tree more than 50% of link field are with

null values thereby wasting memory space.So to

avoid null values in the node we just set the threads

which links to its child or some other node.

The wasted NULL links in storage representationof

binary tree can be replaced by threads

There are 3 ways to thread a binary tree,these

correspond to inorder,preoreder and postorder.

Mostly we use inorder traversal.

Threaded Binary Trees (Continued)

o If a node has no left child, store a pointer to

the node’s inorder predecessor

o If a node has no right child, store a pointer to

the node’s inorder successor

o Here 1st and last node node do not have inorder

predecessor and successor node.To maintain

uniformity of setting of threads we are

maintaining a dummy code,HEADER NODE

HEAD

A

B D

C E G

F

Inorder Traversal

CBAEFDG

HEAD

A

B D

C E G

F

RIGHT THREADED BINARY TREE:-

in this tree the right NULL pointer of each node

points to inorder successor.

Inorder Traversal

CBAEFDG

LEFT THREADED BINARY TREE:-in this tree the

left null pointer of each node points to its

inorder predecessor.

HEAD

A

B D

C E G

F

Inorder Traversal

CBAEFDG

FULL THREADED BINARY TREE:- In this tree

both right & left NULL pointer points to its

inorder successor & inorder predecessor

respectively

HEAD

A

B D

C E G

F

Inorder Traversal

CBAEFDG

A Threaded Binary Tree

A

B C

GE

I

D

H

F

root

inorder traversal:

H, D, I, B, E, A, F, C, G

A FULL Threaded Binary Tree

A

B C

GE

I

D

H

F

root

inorder traversal:

H, D, I, B, E, A, F, C, G

• Storage representation of Threaded Binary tree.

• Here we need two extra variable to differentiate between

thread and links.

• Te variable lthread & rthread are used to indicate whether

left and right pointers are normal pointers or thread.

• Lthread(1) =left normal pointer

• Lthread(0) = left thread

• rthread(1) =right normal pointer

• rthread(0) = right thread

HEAD

A

B D

C E G

F

11

0

00 0

0 0

0 0

1 1 1

1

ADVANTAGE OF THREADED BINARY TREE:-

Traversal operation is faster

We can efficiently determine predecessor &

successor nodes starting from any node

Any node can be accessed from any other node.

Threads are usually upward whereas links are

downward. Thus in a threaded node one can

move either direction.

EXPRESSION TREE:-

It’s a binary tree which stores an arithmetic

expression.

The leaves of expression tree are operands & all

internal nodes are the operators.

(A + B * C) – ( ( D * E + F ) / G )

-

+ /

*A

CB

+ G

*

DE

F

C

B

A

*

B CA

Construction of an expression tree:-

Use postfix notation to construct expression tree.

(A + B * C) – ( ( D * E + F ) / G )

Postfix:- ABC * + DE * F + G / -

*

B C

A

+

(1)(2) (3)

Balanced and unbalanced BST

4

2 5

1 3

1

5

2

4

3

7

6

4

2 6

5 71 3

Is this “balanced”?

Balancing Binary Search Trees

Strict balance

The tree must always be balanced perfectly

Pretty good balance

Only allow a little out of balance

Adjust on access

Self-adjusting

Many algorithms exist for keeping binary

search trees balanced

Adelson-Velskii and Landis (AVL) trees (height-

balanced trees)

Splay trees and other self-adjusting trees

B-trees and other multiway search trees

AVL - Good but not Perfect

BalanceAVL trees are height-balanced binary search trees

That satisfies following conditions

• For every node, heights of left and right subtree can differ by no more than 1

• The left & right subtree of a tree node are also AVL Tree

• Or A binary search tree is said to be AVL tree if all its

nodes have a balance factors 1,0,or -1.

Balance factor of a node

• height(left subtree) - height(right subtree)

Balancing factor is used to determine whether given BST is balanced or not

Different value of Balancing Factor(BF)

BF=0 if BST is perfectly balanced

BF=1 if height of left subtree is one greater than

height of right subtree.

BF=-1 if height of rigtht subtree is one greater

than height of left subtree.

BF<= -2 and BF>=2 that BST is said to be

unbalanced

In AVL tree the value of BF for a node can be -1,0

or 1.

66

40 80

30 50

+1

0 0

0 0

Balanced BST (AVL tree)

66

40 80

30 50

+2

+1 0

0 0

Un Balanced BST

30 30

00

INSERT A NODE INTO AVL TREE

Insertion of a node may or may not result in

unbalanced tree.

Certain situation where insertion may not result

in unbalanced tree are as follows

Node is inserted as child node of which having

one child node

Node is inserted in subtree of lesser height.

Node is inserted in tree having both subtree

having same height.

Insert and Rotation in AVL Trees Insert operation may cause balance factor to

become 2 or –2 for some node

So after the Insert, go back up to the root node

by node, updating heights

If a new balance factor is 2 or –2, adjust tree by

rotation around the node

Let the node that needs rebalancing be .

There are 4 cases:

Outside Cases (require single rotation) :

1. Insertion into left subtree of left child

2. Insertion into right subtree of right child

Inside Cases (require double rotation) :

3. Insertion into right subtree of left child .

4. Insertion into left subtree of right child.

Insertions in AVL Trees

j

k

X Y

Z

Consider a valid AVL subtree:- LL rotation

AVL Insertion: Outside Case

h

hh

j

k

XY

Z

Inserting into X

destroys the AVL

property at node j


h

h+1 h

j

k

XY

Z

Do a “right rotation”


h

h+1 h

j

k

XY

Z

Do a “right rotation”

Single right rotation

h

h+1 h

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

LEFT ROTATION:-

Insertion into left subtree of left child

consider balanced tree shown in figure.

Insertion of node with value 95 result in unbalanced tree

95

66

4080

70 90

66

4080

70 90

-1

00

0 0

-2

0

0

0

-1

-1

RIGHT ROTATION:-

Insertion into left subtree of left child

Insertion of the node with value 25 results in

unbalanced tree.

66

4080

30 50

66

4080

30 50

25

00

0 0

+1

Case 3:-Unbalance occur due to insertion into right subtree of left child.This involves two rotations,Insert node with value 55.

66

4080

30 50

00

0 0

+1

66

4080

30 50

-10

0-1

+2

55

0

Case 4:-Unbalance occur due to insertion into left subtree of right child.This involves two rotations,Insert node with value 70.

66

4080

75 90

-1

00

0 0

66

4080

75 90

-2

0+1

+10

70

0

B-TREE:-B tree is a specialized multiway tree used to

store the records in a disk.

Here each node can store multiple values and can

point to multiple subtrees.A b-tree of order m

satisfies following conditions.

An m-way search tree all nodes are of degree <=m

Each node contains following attribute

All the elements stored in a node are in ascending

order

All the elements stored in left subtree of Pi are less

than of Pi and All the elements stored in right

subtree of Pi are greater than of Pi

P0 K1 P1 K2 P2 …. Kn PnKi = key values

Pi = pointers to subtree of T

20 40

10 15 25

30

35 45 50

A B tree of Order 3

B tree of order 3 means any node

can store atmost 2 elements

and point to atmost 3 subtree

Create btree of order 4 (any node can store

atmost 3 element and point to atmost 4 subtree)

66,90,40,75,30,35,80,70,20,50,45,55,110,100,

120

66

(1) (2)

(3)

(4)

(5)

(6)

Forest

A forest is a set of n >= 0 disjoint trees

A E G

B C D F H I G

H

I

A

B

C

D

F

E

Forest

TREEDELETE(HEAD,T):-

X is the node marked for deletion

PARENT is pointer which denotes address of

parent of node marked for deletion.

CUR:-denotes address of node to be deleted

1.[intialize]

If LPTR(HEAD )!= HEAD

Then CUR LPTR(HEAD)

PARENT HEAD

D ‘L’

Else write (‘NODE NOT FOUND’)

return

[search for the node marked for deletion]

FOUND false

Repeat while not FOUND and CUR != NULL

If DATA(CUR) = X

Then FOUND true

Else if X < DATA(CUR)

then (branch left)

PARENT CUR

CUR LPTR(CUR)

D ‘L’

else (branch right)

PARENT CUR

CUR RPTR(CUR)

D ‘R’

If FOUND = false

Then write(‘NODE NOT FOUND’)

return

3.[perform indicated deletion and restructure

the tree]

If LPTR(CUR) = NULL

Then (empty left sub tree)

Q RPTR(CUR)

Else if RPTR(CUR)=NULL

then (empty right subtree)

Q LPTR(CUR)

else (check right child for successor)

SUC RPTR(CUR)

if LPTR(SUC) = NULL

then LPTR(SUC)LPTR(CUR)

Q SUC

Else (search for successor of CUR)

PRED RPTR(CUR)

SUC LPTR(PRED)

repeat while LPTR(SUC)!= NULL

PRED SUC

SUC LPTR(PRED)

(connect successor)

LPTR(PRED) RPTR(SUC)

LPTR(SUC)LPTR(CUR)

RPTR(SUC) RPTR(CUR)

Q SUC

(connect parent of X to its replacement)

if D= ‘L’

Then LPTR(PARENT) Q

ELSE RPTR(PARENT) Q

prepared by:mitali sonar (lect. computer dep.)

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