The document discusses various data structures and their operations. It begins with an introduction to linear data structures like arrays, linked lists, stacks and queues. It describes array implementation using sequential storage and linked list implementation using pointers. Common operations on these structures like traversal, insertion, deletion and searching are discussed. The document also covers non-linear data structures like trees and graphs and basic tree traversals. It provides examples of applications of different data structures and concludes with definitions of key terms.

1. B.Tech II Sem „CST‟; Scheme 2020; AY: 2020_21
DATA STRUCTURES
Dr. C. Sreedhar

2. Unit 1: Arrays
 Introduction to Data Structures, Definition,
 Classification of Data Structures, Linear and Non Linear
 Sequential Storage Representation: Arrays,
 Operations on Arrays: Insertion, Deletion, Traversing;
 Applications of arrays: Linear Search, Binary Search,
Bubble Sort, Selection Sort, Insertion Sort, Merging of
arrays

3. Unit 2: Linked List
 Linked Storage Representation: Linked Lists
 Linked storage representation using pointers,
 Types of Linked Lists: Single linked list
 Double linked list
 Operations on linked lists: Traversing
 Searching, Insertion and Deletion

4. Unit III: Stack
 Linear Data Structures: Stacks
 Representation of stack using sequential storage
and linked allocation methods
 Operations on stacks:
 Push, Pop and Display

5. Unit IV: Queue
 Linear Data Structures: Queues
 Representation of Queue using sequential and
linked allocation
 Operations on Queues: Insertion, Deletion and
Traversing
 Circular queue

6. Unit V: Non Linear DS
 Non Linear Data Structures: Trees
 Basic terminology, Binary Trees, Representation of
Binary tree in memory using arrays and linked lists.
 Binary search trees, operations on binary search trees
 Insertion, Deletion
 Recursive traversals: Preorder, Inorder and Postorder

8. Definitions
 Data: a value or set of values
 Entity: is any object in the system that we want to
model and store information about
 Information: Processed data to gain knowledge
 Data type: is a classification of type of data that
a variable or object can hold.
 Built-in Data type

9.  Abstract Data Types (aka ADTs) are descriptions of
how a data type will work without implementation
details.
 Ex: Stack

10. Data Structure: Definition
 Data structure (DS) is representation of the logical
relationship existing between individual elements of data.
 Data structure is a data organization, management and
storage format that enables efficient access and
modification.
 Data structures provide a means to manage large amounts
of data efficiently for uses such as large databases
 Program=algorithm + Data Structure

11. DS
Linear DS Non-Linear DS
Array
Linked List Stack
Queue Graph Trees
DS : Classification

12. Linear Vs. Non-linear Data Structure
 The basic difference between linear and non-linear
data structure is that in linear data structures, for
each element there is a fixed next element but in
case of non-linear data structure, each element can
have many different next elelemts.

13. Linear DS
 Array: collection of finite number of homogeneous data elements
 Linked List: data elements are managed by collection of nodes,
where each node contains link or pointer which points to next
node in the list.
 Stacks: linear collection of data elements in which insertion and
deletion are restricted at only one end. (LIFO)
 Queues: linear collection of data elements in which insertion can
take place at one end and deletion can take place at other end.
(FIFO)

14. Identify the Data Structure
Array Stack
Queue

15. Stack
Queue
Linked
List

16. Non-Linear Data Structures
 Trees: is a non-linear data structure which is used to
represent data elements having hierarchical
relationship between them
 Graphs: is a non-linear data structure which is used to
represent data having relationship among its elements
which are not necessarily hierarchical in nature.
 Set: allows to add data to a container
 Table

17. Identify non-linear Data Structure
Tree Table Graph
Set

18. Graph

20. Applications of DS
 Stack: functions and stack, expression evaluation,
UNDO/REDO operation in word processors etc.
 Queues: Operating systems often maintain a queue of
processes that are ready to execute or that are waiting for
a particular event to occur.
 Graphs - Connections/relations in social networking sites,
Routing ,networks of communication, data organization etc.

21. Applications of Data Structures
 Dictionary: Arranged and ordered and unordered words
 Android Auto : City map, source, destination, shortest path
 Student details: Roll no, Name, age, address, marks, ...
 Library: Arranging books in certain order
 Travelling Salesperson problem
 Airport: Flights(Source, destination, connecting), companies...

22. Applications: Trees

23. largest element of an array
for(i = 1; i < n; i++)
{
if(arr[0] < arr[i])
arr[0] = arr[i];
}
printf("Largest element = %.2f", arr[0]);

24. Solved Example: Accept elements of array and display largest element
int i, n; float arr[100];
// Accept the value of n
for(i = 0; i < n; ++i)
{
printf("Enter Number %d: ", i+1); scanf("%f", &arr[i]);
}
for(i = 1; i < n; ++i)
{
if(arr[0] < arr[i])
arr[0] = arr[i];
}
printf("Largest element = %.2f", arr[0]);

25. Sequential Storage Representation
 Arrays:
 One dimensional Arrays
 Declaration; initialization; passing arguments to a function
 Two dimensional Arrays
 Declaration; initialization; passing arguments to a function
 Arrays Storage representation: address calculation

26. 1 D Array: Finding address
 First location of an array is called Base Address

27. 1D Array: Finding address
 To find the address of element of an array, say A[i],
 Address of A [ i ] = B + W * ( i – LB ); where
 B = Base address
W = Storage Size of one element stored in array
i = Subscript of element whose address is to be found
LB = Lower limit or Lower Bound of subscript


28. Finding address: 1D Array: Example 1
 Given Array A[20]; ie., A[0], A[1],......A[19]
 Consider B: 2000; W= 4 bytes; i= 12; Find the
address of where A[12] is stored?
 Address of A [ i ] = B + W * ( i – LB )
 A [12] = 2000 + 4 * (12-0)
= 2048
 Ans: The address of A[12] = 2048

29. Address calculation in 2D Arrays
 Given X[2][2]
Two types of arrangement
 Row major
 Column major

30. Row Major System:
Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ]
Column Major System:
Address of A [ I ][ J ] = B + W * [( I – Lr ) + M * ( J – Lc )]
Where,
B = Base address
I = Row subscript of element whose address is to be found
J = Column subscript of element whose address is to be found
W = Storage Size of one element stored in the array (in byte)
Lr = Lower limit of row/start row index of matrix, if not given assume 0 (zero)
Lc = Lower limit of column/start column index of matrix, if not given assume 0 (zero)
M = Number of row of the given matrix
N = Number of column of the given matrix

31. 2D Array Address Calculation(Row Major)
 Consider A[3][4]; with base address 2000. Find the
address of A[1][2] using row major.
 Address of A [ I ][ J ] = B + W * [N*(I – Lr)+(J–Lc) ]
 Given:
 I = 1; J = 2; W = 4 bytes; B = 2000;
 N: Total no. of cols = 4; Lr: 0; Lc: 0
 A[1][2]=2000+4*[4*(1-0)+(2-0)] = 2024

33. 2D Array Address Calculation(Col Major)
 Consider A[3][4]; with base address 2000. Find the
address of A[1][2] using Column major.
 Address of A [ I ][ J ] = B+W*[(I – Lr)+M*( J – Lc )]
 Given:
 I = 1; J = 2; W = 4 bytes; B = 2000;
 M: Total no. of rows = 3; Lr: 0; Lc: 0
 A[1][2]=2000+4*[(1-0)+3*(2-0)] = 2028

34. Column Major

35. 2D Array: Find address(Row Major)
 Consider A[10][20]; with base address 501. Find
the address of A[8][10] using row major.
 Address of A [ I ][ J ] = B + W * [N*(I – Lr)+(J–Lc) ]
 Given:
 I = 8; J = 10; W = 4 bytes; B = 501;
 N: Total no. of cols = 20; Lr: 0; Lc: 0
 A[8][10]=501+4*[20*(8-0)+(10-0)] = 1181

36. 2D Array: Find address(Col Major)
 Consider A[10][20]; with base address 501. Find
the address of A[8][10] using Column major.
 Address of A [ I ][ J ] = B+W*[(I – Lr)+M*( J – Lc )]
 Given:
 I = 8; J = 10; W = 4 bytes; B = 501;
 M: Total no. of rows = 10; Lr: 0; Lc: 0
 A[8][10]=501+4*[(8-0)+10*(10-0)] = 933

37. largest element of an array
for(i = 1; i < n; i++)
{
if(arr[0] < arr[i])
arr[0] = arr[i];
}
printf("Largest element = %.2f", arr[0]);

38. Delete element in an Array
main()
{
....
AcceptArray(n,a);
DisplayArray(n,a);
Insert(n,a);
DisplayArray(n,a);
Delete(n,a);
DisplayArray(n,a);
}
1. Accept the position
2. Check whether position > = n+1;
True: Deletion not possible
False:
a) for(i=pos-1;i<n;i++)
a[i]=a[i+1]
b) n - -
c) print the elements of an array
3. Stop

39. Linear Search (Single Occurrence)
SEARCH (A [ ], n, s)
1. for(i=0; i<n; i++)
1a. if (s=A[i]) break;
2. if (i<n)
2a. Print "Element found at index, i"
3. else Print "Element not found "
4. STOP

40. Linear Search (Multiple occurrences)
SEARCH (A [ ], n, s)
1. Initialize count = 0
1. for(i=0; i<n; i++)
1a. if (A[i]=s)
1a.i. Print "element is found at location (i+1)"
1a.ii. count++
2. if (count=0)
2a. Print "Element is not found "
3. else Print "Element is found at count times in array "
4. STOP

41.  void read(int [ ],int);
 void display(int [ ],int);
 void search(int [ ],int,int);
main( ) {
//Accept n
read(a,n);
display(a,n);
do {
printf("Enter searching element:");
// Accept and store in „se‟ var
search(a,n,se);
printf("search another element?y/n: ");
scanf(" %c",&ch);
} while(ch=='y');
void read(int a[ ],int n)
{ int i; for(i=0;i<n;i++) scanf("%d",&a[i]);
}

42. void read(int a[ ],int n) { int i; for(i=0;i<n;i++) scanf("%d",&a[i]); }
void display(int a[ ],int n) { int i; for(i=0;i<n;i++) printf("%d ",a[i]);}
// Search for multiple occurrences
void search(int a[ ], int n, int se)
{
int i,count=0;
for(i=0;i<n;i++)
{ if(a[i]==se) { printf("%d is found at %d",se,i+1); count++; } }
if(count==0) printf("n%d is not found",se);
else printf("n%d is present %d times in array",se,count);
}

43. Binary Search: Algorithm
1. Initialize lb=0, ub=n-1, count = 0
2. while (lb < = ub)
2a. mid=(lb+ub)/2;
2b. if(se==a[mid])
2b.i. Print "element is found
at position,se,mid+1
2b.ii. count++;
2b.iii. break;
2c. else if(se<a[mid])
2c.i. ub=mid-1;
2d. else
2d.i lb=mid+1;
3. End while
4. if(count==0)
4a. print "Element not found"
5. STOP

44. Merging two arrays
void read(int [ ],int);
void display(int [ ],int);
void merge(int [ ],int [ ],int [ ],int,int);
main()
//Accept n
Read(a,n)
read(b,n);
display(a,m);
display(b,n);
merge(a,b,c,m,n);
display(c,m+n);

45. void read(int a[ ],int m) { int i; for(i=0;i<m;i++) scanf("%d",&a[i]); }
void display(int a[ ],int m) { int i; for(i=0;i<m;i++) printf("%d ",a[i]); }
void merge(int a[ ],int b[ ],int c[ ],int m,int n)
{
int i,j,k;
i=j=k=0;
while(i<m&&j<n)
{
if(a[i]<b[j]) c[k++]=a[i++];
else c[k++]=b[j++];
}
while(i<m) c[k++]=a[i++];
while(j<n) c[k++]=b[j++];
}

46. Finding Largest no. in array using fn‟s
 void CreateArray(int n,int a[ ]);
 void DisplayArray(int n,int a[ ]);
 void Largest(int n,int a[ ]);
void CreateArray(int n,int a[ ]) { int i; for(i=0;i<n;i++)
{ printf("Enter a[%d]:",i); scanf("%d",&a[i]); } }
int main()
{
.........
CreateArray(n,a);
DisplayArray(n,a);
Largest(n,a);
.....
}
void DisplayArray(int n,int a [ ]) { int i; for(i=0;i<n;i++) {
printf("%dt",a[i]); } printf("n"); }
void Largest(int n,int a[ ]) { int i; for(i=1;i<n;i++) {
if(a[0]<a[i]) a[0]=a[i]; }
printf("Largest element = %d",a[0]);
}

47. Lab1a) Array op‟s: insert,delete,traverse
void create();
void display();
void insert();
void del();
void create() { int i; for(i=0;i<n;i++) {
printf("Enter a[%d]:",i); scanf("%d",&a[i]); } }
int main()
{
.........
do { create(); display()
insert(); delete(); exit();
}while(..)
.....
}
void display() { int i; for(i=0;i<n;i++) { printf("%dt",a[i]); } }
void insert( ) {
// Accept the position to insert and store in „pos‟ variable
// Accept the value to insert and store in „val‟ variable
for(i=n-1;i>=pos-1;i--) a[i+1]=a[i]; a[pos-1]=val; n=n+1; }
void del() { // Accept the „pos‟; val=a[pos-1];
for(i=pos-1;i<n;i++) a[i]=a[i+1]; n=n-1;
printf("nThe deleted element is =%d",val); }

48. Lab 2a) Linear Search
 void read(int [ ],int);
 void display(int [ ],int);
 void search(int [ ],int,int);
void search(int a[ ], int n, int s) { // Single Occurance
...
for(i=0; i<n; i++) if (s=A[i]) break;
if (i<n) Print "Element found at index, i“
else Print "Element not found “ }
main() { //Accept n
read(a,n); display(a,n);
do { // Accept „se‟
search(a,n,se); }while(..);
}
void search (int a [ ], int n, int s){ // Multiple Occurances ...
Initialize count = 0
for(i=0; i<n; i++) if (A[i]=s)
{ Print "element is found at location (i+1)“
count++ } if (count=0) Print "Element is not found “
else Print "Element is found at count times in array “
}

49. Lab 2b) Binary Search
 void read(int [ ],int);
 void display(int [ ],int);
 void search(int [ ],int,int);
main() { //Accept n
read(a,n); display(a,n);
do { // Accept „se‟
search(a,n,se); }while(..);
}
void search(int a[ ],int n,int se) {
int lb,ub,mid,count=0; lb=0; ub=n-1;
while(lb<=ub) { mid=(lb+ub)/2;
if(se==a[mid]) { printf("nThe element %d is found
at position %dn",se,mid+1); count++; break;
}
else if(se<a[mid]) ub=mid-1;
else lb=mid+1;
}
if(count==0) printf("nElement not found");
}

50. Binary Search: Algorithm
1. Initialize lb=0, ub=n-1, count = 0
2. while (lb < = ub)
2a. mid=(lb+ub)/2;
2b. if(se==a[mid])
2b.i. Print "element is found
at position,se,mid+1
2b.ii. count++;
2b.iii. break;
2c. else if(se<a[mid])
2c.i. ub=mid-1;
2d. else
2d.i lb=mid+1;
3. End while
4. if(count==0)
4a. print "Element not found"
5. STOP

51. Merging of two sorted arrays
 Enter size of first array: 4
 Enter elements of first array: 1 3 5 7
 Enter size of Second array: 4
 Enter elements of Second array: 2 4 6 8
 Merged Final Sorted array elements are:
1 2 3 4 5 6 7 8

52. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A B
C

53. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A B
C
i j
1 2
<
1

54. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
j
1 2
<
1
i
3
2

55. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
1 2
<
1
i
3
2
j
4
3

56. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
1 2
<
1
i
3
2
j
4
3
5
4

57. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
1 2
<
1
i
3
2
j
4
3
5
4
6
5

58. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
1 2
<
1
i
3
2
j
4
3
5
4
6
5
7
6

59. Merging of two sorted arrays
1 3 5 7 2 4 6 8
A
B
C
1 2
<
1
i
3
2
j
4
3
5
4
6
5
7
6
8
7 8

60. How to code?
if ( a [ i ] < b [ j ] )
c[ k++ ] = a[ i++ ];
else
c[ k++ ] = b[ j++ ];
while( i<m && j<n )
{
}
while( i < m ) c[ k++ ] = a[ i++ ];
while( j < n ) c[ k++ ] = b[ j++ ];

61. Merging function definition
void merge( ..........)
{
int i,j,k; i=j=k=0;
while( i<m && j<n )
{
if ( a [ i ] < b [ j ] ) c[ k++ ] = a[ i++ ];
else c[ k++ ] = b[ j++ ];
}
while( i < m ) c[ k++ ] = a[ i++ ];
while( j < n ) c[ k++ ] = b[ j++ ];

62. Merging two arrays: Complete Program
void read(int [ ],int);
void display(int [ ],int);
void merge(int [ ],int [ ],int [ ],int,int);
main() {
//Accept n
read(a,n)
read(b,n);
display(a,m);
display(b,n);
merge(a,b,c,m,n);
display(c,m+n); ..}

63. void read(int a[ ],int m) { int i; for(i=0;i<m;i++) scanf("%d",&a[i]); }
void display(int a[ ],int m) { int i; for(i=0;i<m;i++) printf("%d ",a[i]); }
void merge(int a[ ],int b[ ],int c[ ],int m,int n) {
int i,j,k; i=j=k=0;
while(i<m&&j<n) {
if(a[i]<b[j]) c[k++]=a[i++];
else c[k++]=b[j++];
}
while(i<m) c[k++]=a[i++];
while(j<n) c[k++]=b[j++];
}

64. Sorting
 Sorting is the process of arranging unordered elements
into ordered (Ascending or descending order) elements.
 Sorting algorithms:
 Bubble Sort; Selection Sort; Recursive Bubble Sort;
 Insertion Sort; Merge Sort; Quick Sort; Heap Sort
 Radix sort; Bucket Sort; Shell Sort; Pigeonhole Sort;
 Pancake Sort; Gnome Sort; Stooge Sort; Tag Sort; Tree Sort

65. Bubble Sort: Input and Output
 Enter the size of the array : 6
 Enter the elements of the array : 4 3 6 1 5 2
 After sorting, elements of the array are:
1 2 3 4 5 6

66. Bubble Sort: Logic
6, 2, 9, 11, 9, 3, 7, 12
i
j
a[ j ] a[ j+1 ]
> then swap < then increment j
Swap
temp = a[ j ]
a[ j ] = a[ j+1 ]
a[ j+1 ] = temp
Iterate range of j: 0 to j < n-1-i
Iterate range of i: 0 to i < n-1

67. Bubble Sort: How to Code?
if(a[ j ] > a[ j+1 ])
{
temp = a[ j ];
a[ j ] = a[ j+1 ];
a[ j+1 ] = temp;
}
for(i=0;i < n-1;i++)
{
for(j=0;j < n-1-i;j++)
{
}
}

68. Defining Bubble Sort function
void sort(int a[ ],int n)
{
int i,j,temp;
for(i=0;i<n-1;i++)
{
for(j=0;j<n-1-i;j++)
{ if(a[ j ]>a[ j+1 ]) {temp=a[j]; a[j]=a[j+1]; a[j+1]=temp;}
}
}
}

69. Bubble Sort: Program
void read(int [ ],int);
void display(int [ ],int);
void bsort(int [ ],int);
int main() { // Declare necessary variables
// Accept size of the array and store in „n‟ variable
do
{
read(a,n);
display(a,n);
bsort(a,n);
printf(“Array after sorting are:");
display(a,n);
// Accept ch and decide to repeat or exit
} while(ch=='y'); return 0; }
// Define read(int [ ],int)
// Define display(int [ ],int)
// Define bsort(int [ ],int)

70. Selection Sort
void ssort ( )
{
int i, j, min, temp;
for (i=0; i<n; i++)
{
min = i;
for ( j=i+1; j<n; j++)
{
if (a[ j ] < a[min]) min = j;
}
temp = a[i]; a[i] = a[min]; a[min] = temp;
}
}

71. Selection Sort: Complete Program
void ssort ( );
int a[30], n;
int main()
{
// Declare variables
// Accept the size of array, store in „n‟
// Accept the elements of array, store in a[i]
ssort();
printf("nnAfter sorting:n");
// Display elements of array
}
void ssort ( )
{
int i, j, min, temp;
for (i=0; i<n; i++)
{
min = i;
for ( j=i+1; j<n; j++)
{
if (a[ j ] < a[min]) min = j;
}
temp = a[i]; a[i] = a[min]; a[min] = temp;
}
}

72. Insertion Sort: Logic

73. 5 4 3
2 6 1
5 4 3
2 6 1
2 5
4 5 j=0
6
1
1 5
4
1
2
1 3 6
3 5
3 4 j=1
j=2
j=3
j=4
j=5

74. Insertion Sort
void isort(int a[ ],int n)
{
int i,j,key;
for(i=1;i<n;i++)
{
key=a[i];
for( j = i-1; j >= 0 && a[ j ] > key; j--) { a[ j+1]=a[ j ]; }
a[j+1]=key;
}
}

75. Insertion Sort: Complete program
void read(int [ ],int);
void display(int [ ],int);
void isort(int [ ],int);
int main()
{
// Declare necessary variables
// Accept size and store it in „n‟ variable
read(a,n);
display(a,n); // Before sorting
isort(a,n);
display(a,n); // After sorting
return 0;
}
// Define read(int [ ],int);
// Define display(int [ ],int);
// Define isort(int [ ],int);

76. Merge sort example
index 0 1 2 3 4 5 6 7
value 22 18 12 -4 58 7 31 42
22 18 12 -4
22 18
22 18
18 22
merge
split
12 -4
12 -4
-4 12
merge
split
split
-4 12 18 22
58 7 31 42
58 7
58 7
7 58
merge
split
31 42
31 42
31 42
merge
split
split
7 31 42 58
-4 7 12 18 22 31 42 58
split
merge merge
merge

77. Merge Sort: Function
void sort(int a[ ],int lb,int ub)
{
int mid;
if(lb<ub)
{
mid=(lb+ub)/2;
sort(a,lb,mid);
sort(a,mid+1,ub);
merge(a,lb,mid,ub);
}
}
void merge(int a[ ],int lb,int mid,int ub)
{
int b[50],i,j,k;
i=lb,j=mid+1,k=lb;
while(i<=mid&&j<=ub)
{
if(a[i]<a[j]) b[k++]=a[i++]; else b[k++]=a[j++];
}
while(i<=mid) b[k++]=a[i++];
while(j<=ub) b[k++]=a[j++];
for(i=lb;i<=ub;i++) a[i]=b[i];
}

78. Merge Sort: Complete Program
void read(int [ ],int);
void display(int [ ],int);
void sort(int [ ],int,int);
void merge(int [ ],int,int,int);
int main()
{
// Declare necessary variables
// Accept size of array and store it in „n‟
read(a,n);
display(a,n); // Before sorting
sort(a,0,n-1);
display(a,n); // After Sorting
return 0;
}
// Define read(int [ ],int)
// Define display(int [ ],int)
// Define sort(int [ ],int,int)
//Define merge(int [ ],int,int,int)

79. Unit 2: Linked Lists
 Linked Storage representation using pointers
 Types of Linked List:
 Single Linked List
 Double Linked List
 Operations on Linked Lists:
 Traversing
 Searching
 Insertion and Deletion

80. Linked List
 A Linked list is an ordered collection of finite,
homogeneous data elements called nodes, where
the linear order is maintained by means of links or
pointers.
 Each node is composed of data and a reference to
the next node in the sequence. This structure allows
for efficient insertion or removal of elements from
any position in the sequence during iteration.

81. Advantages of Linked Lists
 Linked lists are a dynamic data structure, which can grow and shrink during
the execution of the program.
 Insertion and deletion node operations are easily implemented in a linked list.
 Dynamic data structures such as stacks, queues can be implemented using
linked list.
 There is no need to define an initial size for a linked list.
 Items can be added or removed from the middle of list.
 Linked lists have efficient memory utilization. Memory is not pre-allocated in
linked list. Memory is allocated whenever it is required and is de-allocated
when it is no longer needed.

82. Disadvantages of Linked Lists
 They use more memory than arrays because of the storage
used by their pointers.
 Nodes in a linked list must be read in order from the beginning
as linked lists are inherently sequential access.
 Difficulties arise in linked lists when it comes to reverse
traversing. For instance, singly linked lists are cumbersome to
navigate backwards and while doubly linked lists are
somewhat easier to read, memory is consumed in allocating
space for a back-pointer.

83. Types of LL
 Single Linked List
 Double Linked List
 Circular Linked List

84. Create SLL
head
Data *link
one
Create a node One,
which contains data
and a link. Assign initial
value of data to 10
and link to NULL.
Traverse the single
linked list

85. SLL
typedef struct sll
{
int data;
struct sll *next;
}node;
node *head=NULL;
void display( );
void main()
{
int n;
node *one; one=NULL;
one=(node *) malloc (sizeof (node) );
onedata=10;
one  next=NULL;
head=one; display( );
}

86. Traversing SLL
void display()
{
node *temp;
if(head==NULL) printf("List is empty");
else
{
temp=head;
while(temp!=NULL)
{
printf(“n%d",temp->data);
temp=temp->next;
}
}
}

87. Create SLL with the following
head
Data(10) *next
one two
Data(20) *next Data(30) NULL
three

88. Coding SLL
typedef struct sll
{
int data;
struct sll *next;
}node;
node *head=NULL;
void main()
{
int n;
node *one,*two,*three;
one=NULL; two=NULL; three=NULL;
one=(node *)malloc(sizeof(node));
two=(node *)malloc(sizeof(node));
three=(node *)malloc(sizeof(node));
onedata=10; two  data=20; three  data=30;
one->next=two; two->next=three; three->next=NULL;
head=one;
display();
getch();
}

89. Create n nodes in SLL
typedef struct sll
{ int data; struct sll *next; }node;
node *head=NULL;
void createList(int n);
void traverseList();
int main()
{
int n;
printf("Enter n: "); scanf("%d", &n);
createList(n);
traverseList();
return 0;
}
void createList(int n)
{
node *newNode,*temp; int data, i;
head = (node *)malloc(sizeof(node));
if(head == NULL) printf("Unable to allocate memory.");
else
{
printf("Enter the data of node 1: "); scanf("%d", &data);
head->data = data; head->next = NULL;
temp = head;
for(i=2; i<=n; i++)
{
newNode = (node *)malloc(sizeof(node));
if(newNode == NULL)
{ printf("Unable to allocate memory.");
break; }
else
{
printf("nEnter data of node %d: ", i); scanf("%d", &data);
newNode->data = data; newNode->next = NULL;
temp->next = newNode;
temp = temp->next; } } } }

90. Single Linked List Creating node
typedef struct slist
{
int no;
char name[50];
struct slist *link;
} node;
node *first=NULL;
no name *link
first
node
NULL

91. Linked Lists
 A linked list is a series of connected nodes
 Each node contains at least
 A piece of data (any type)
 Pointer to the next node in the list
 Head: pointer to the first node
 The last node points to NULL
A 
Head
B C
A
data pointer
node

92. Create SLL with the following
head
Data(10) *next
one two
Data(20) *next Data(30) NULL
three

93. SLL: Create three nodes and traverse
typedef struct sll
{
int data;
struct sll *next;
}node;
node *head=NULL;
void main()
{
int n;
node *one,*two,*three;
one=NULL; two=NULL; three=NULL;
one=(node *)malloc(sizeof(node));
two=(node *)malloc(sizeof(node));
three=(node *)malloc(sizeof(node));
onedata=10; two  data=20; three  data=30;
one->next=two; two->next=three; three->next=NULL;
head=one;
display();
}

94. Create n nodes in SLL
typedef struct sll
{ int data; struct sll *next; }node;
node *head=NULL;
void createList(int n);
void traverseList();
int main()
{
int n;
printf("Enter n: "); scanf("%d", &n);
createList(n);
traverseList();
return 0;
}
void createList(int n)
{
node *newNode,*temp; int data, i;
head = (node *)malloc(sizeof(node));
if(head == NULL) printf("Unable to allocate memory.");
else
{
printf("Enter the data of node 1: "); scanf("%d", &data);
head->data = data; head->next = NULL;
temp = head;
for(i=2; i<=n; i++)
{
newNode = (node *)malloc(sizeof(node));
if(newNode == NULL)
{ printf("Unable to allocate memory."); break; }
else
{
printf("nEnter data of node %d: ", i); scanf("%d", &data);
newNode->data = data; newNode->next = NULL;
temp->next = newNode;
temp = temp->next; } } } }

95. Insert node at the beginning
48 17 142
head //
head 93
Step 1 Step 2
Step 3

96. Insert node at the end of the list
48 17 142
head //
Step 1 Step 2
Step 3

97. Insert node at specified position
Step 1 Step 2
Step 3
Step 4

98. SLL: Inserting a node
void insertion( )
{
node *p,*save,*ptr;
p=(node *) malloc(sizeof(node));
printf("nEnter rollno and name:");
scanf("%d %s",&pno,pname);
if(first==NULL)
{
plink=NULL;
first=p;
return;
}
save
ptr
1001 sree *link
first
NULL
p
p
1001 sree NULL

99. SLL: Insertion
if(p->no <= first->no)
{
p->link = first;
first = p;
return;
}
save = first; ptr = firstlink;
while(ptr !=NULL && ptrno < pno)
{
save = ptr;
ptr = ptr->link;
}
savelink = p;
plink = ptr;
}
1002 sri *link
p
1002 < = 1001
False
first
save
ptr
1001 sree NULL

100. SLL: Insertion
if(p->no <= first->no)
{
p->link = first;
first = p;
return;
}
save = first; ptr = firstlink;
while(ptr !=NULL && ptrno < pno)
{
save = ptr;
ptr = ptr->link;
}
savelink = p;
plink = ptr;
}
1004 dennis *link
p
1004 < = 1001
False
first
save
ptr
1001 sree NULL

101. void createList(int n);
void insertB(int data);
void insertE(int data);
void insertM(int data, int position);
void deleteF();
void displayList();
printf("nEnter data to insert at beginning of the list: ");
scanf("%d", &data);
insertB(data);
displayList();
printf("nEnter data to insert at middle of the list: ");
scanf("%d", &data);
printf("Enter the position to insert new node: " );
scanf("%d", &position);
insertM(data, position);
displayList();
printf("nEnter data to insert at end of the list: ");
scanf("%d", &data);
insertE(data);
printf("nPress 1 to delete first node: ");
scanf("%d", &choice);
if(choice == 1) deleteF();
displayList();

102. void insertB(int data)
{
struct node *newNode;
newNode = (struct node*)malloc(sizeof(struct node));
if(newNode == NULL) { printf("Unable to allocate memory."); }
else
{
newNode->data = data; // Link data part
newNode->next = head; // Link address part
head = newNode; // Make newNode as first node
printf("DATA INSERTED SUCCESSFULLYn");
}
}

103. void insertM(int data, int position)
{
int i; struct node *newNode, *temp;
newNode = (struct node*)malloc(sizeof(struct node));
if(newNode == NULL) { printf("Unable to allocate memory."); }
else
{
newNode->data = data; // Link data part
newNode->next = NULL;
temp = head;
for(i=2; i<=position-1; i++) { temp = temp->next; if(temp == NULL) break;}
if(temp != NULL)
{
newNode->next = temp->next;
temp->next = newNode;
printf("DATA INSERTED SUCCESSFULLYn");
}
else printf("UNABLE TO INSERT DATA AT THE GIVEN POSITIONn"); } }

104. void insertB(int data)
{
struct node *newNode;
newNode = (struct node*)malloc(sizeof(struct node));
if(newNode == NULL)
printf("Unable to allocate memory.");
else
{
newNode->data = data;
newNode->next = head;
head = newNode;
printf("DATA INSERTED SUCCESSFULLYn");
}
}

105. void insertM(int data, int position)
{
int i;
struct node *newNode, *temp;
newNode = (struct node*)malloc(sizeof(struct node));
if(newNode == NULL)
printf("Unable to allocate memory.");
else
{
newNode->data = data; // Link data part
newNode->next = NULL;
temp = head;

106. for(i=2; i<=position-1; i++)
{
temp = temp->next;
if(temp == NULL)
break;
}
if(temp != NULL)
{
/* Link address part of new node */
newNode->next = temp->next;
/* Link address part of n-1 node */
temp->next = newNode;
printf("DATA INSERTED SUCCESSFULLYn");
}
else
{
printf("UNABLE TO INSERT DATA AT THE GIVEN POSITIONn");
}
}
}

107. void insertE(int data)
{
struct node *newNode, *temp;
newNode = (struct node*)malloc(sizeof(struct node));
if(newNode == NULL)
{
printf("Unable to allocate memory.");
}
else
{
newNode->data = data;
newNode->next = NULL;
temp = head;
while(temp->next != NULL)
temp = temp->next;
temp->next = newNode; // Link address part
printf("DATA INSERTED SUCCESSFULLYn");
}
}

108. void deleteF()
{
struct node *toDelete;
if(head == NULL)
printf("List is already empty.");
else
{
toDelete = head;
head = head->next;
printf("nData deleted = %dn", toDelete->data);
/* Clears the memory occupied by first node*/
free(toDelete);
printf("SUCCESSFULLY DELETED FIRST NODE FROM LISTn");
}
}

109. Double Linked List
struct node { int data; struct node *prev, *next; };
struct node *head = NULL;
static int node_count = 0;
struct node *createNode(int);
void insertNode(int);
void deleteNode(int);
void display();

110. /* creating a new node */
struct node* createNode(int data)
{
struct node *ptr = (struct node *)malloc(sizeof (struct node));
ptr->data = data;
ptr->prev = NULL; ptr->next = NULL;
return (ptr);
}
void display() {
struct node *ptr = head; int i = 0;
while (ptr) { printf("data in node%d:%dn", i, ptr->data);
ptr = ptr->next; i++;
}
}

111. /* Insert a new node at the end of the list */
void insertNode(int data)
{
struct node *temp, *ptr = createNode(data);
if (!head) { head = ptr; node_count++; return; }
else { temp = head;
while (temp) {
if (temp->next == NULL)
{ temp->next = ptr; ptr->prev = temp;
node_count++; return; }
temp=temp->next;
} } }

112. /* Insert a new node at the end of the list */
void insertNode(int data)
{
struct node *temp, *ptr = createNode(data);
if (!head) { head = ptr; node_count++; return; }
else { temp = head;
while (temp) {
if (temp->next == NULL)
{ temp->next = ptr; ptr->prev = temp;
node_count++; return; }
temp=temp->next;
} } }

113. /* Insert a new node at the end of the list */
void insertNode(int data)
{
struct node *temp, *ptr = createNode(data);
if (!head) { head = ptr; node_count++; return; }
else { temp = head;
while (temp) {
if (temp->next == NULL)
{ temp->next = ptr; ptr->prev = temp;
node_count++; return; }
temp=temp->next;
} } }

114. Double Linked List: Deletion
head(200) (400)
NULL 10 (400) (200) 20 (600)
(600)
(400) 30 (800)
(800)
(600) 40 NULL
delete(10)
head(400)
NULL 20 (600)
(600)
(400) 30 (800)
(800)
(600) 40 NULL

115. Double Linked List: Deletion
head(200) (400)
NULL 10 (400) (200) 20 (600)
(600)
(400) 30 (800)
(800)
(600) 40 NULL
delete(20)
head(200)
NULL 10 (600)
(600)
(200) 30 (800)
(800)
(600) 40 NULL

116. Double Linked List: Deletion
head(200) (400)
NULL 10 (400) (200) 20 (600)
(600)
(400) 30 (800)
(800)
(600) 40 NULL
delete(30)
head(200)
NULL 10 (400)
(400)
(200) 20 (800)
(800)
(400) 40 NULL

117. Double Linked List: Deletion
head(200) (400)
NULL 10 (400) (200) 20 (600)
(600)
(400) 30 (800)
(800)
(600) 40 NULL
delete(40)
head(200)
NULL 10 (400)
(400)
(200) 20 (800)
(600)
(400) 30 NULL

118. /* delete the node with given data */
void deleteNode(int data)
{
struct node *ptr, *temp = head;
if (head == NULL) { printf("Data unavailablen"); return; }
else if (temp->data == data)
{
ptr = temp->next; temp->next = NULL; free(temp);
head = ptr; node_count--;
}
else {
while (temp->next != NULL && temp->data != data)
{ ptr = temp; temp = temp->next; }
if (temp->next == NULL && temp->data != data)
{ printf("Given data unvavailable in listn"); return; }

119. /* delete the node with given data */
void deleteNode(int data)
{
struct node *ptr, *temp = head;
if (head == NULL) { printf("Data unavailablen"); return; }
else if (temp->data == data)
{
ptr = temp->next; temp->next = NULL; free(temp);
head = ptr; node_count--;
}
else {
while (temp->next != NULL && temp->data != data)
{ ptr = temp; temp = temp->next; }
if (temp->next == NULL && temp->data != data)
{ printf("Given data unvavailable in listn"); return; }

120. else if (temp->next != NULL && temp->data == data)
{
ptr->next = temp->next;
temp->next->prev = temp->prev;
temp->next = NULL;
temp->prev = NULL;
free(temp);
printf("Data deleted successfullyn"); node_count--;
}
else if (temp->next == NULL && temp->data == data)
{
ptr->next = NULL; temp->next = temp->prev = NULL;
free(temp); printf("Data deleted successfullyn"); node_count--;
}
}
}

121. int main() {
int ch, data;
while (1)
{
printf("1.Insertionn2.Deletion 3. Display 4.Exitn Enter your choice n");
scanf("%d", &ch);
switch (ch)
{
case 1: printf("Enter data to insert:"); scanf("%d", &data); insertNode(data); break;
case 2: printf("Enter data to delete:"); scanf("%d", &data); deleteNode(data); break;
case 3: display(); break;
case 4: exit(0);
default: printf("please enter right optionn"); break;
}
}
}

122. Stack
Stack is an ordered list of similar data type.
Stack is a linear data structure in which the insertion and deletion
operations are performed at only one end.
Stacks can be implemented using either array or LL.
Stack follows LIFO (Last In First Out) principle.
push() function is used to insert new elements into Stack and pop()
function is used to remove an element from stack.
Both insertion and removal are allowed at only one end of Stack
called Top.
Stack is said to be in Overflow state when it is completely full
and is said to be in Underflow state if it is completely empty.

123. Applications of Stacks
 Many compilers use a stack for parsing the syntax of
expressions, program blocks etc. before translating into
low level code.
 Stack is used to evaluate prefix, postfix and infix
expressions.
 Reverse a string
 Processing function calls and return addresses

124. Stack Operations
 Push an element into stack
 Pop an element from stack
 Display the elements of stack
 Count the number of elements in the stack

125. #define MAX 6
int top=-1,stack[MAX];
void push(); void pop(); void display();
int main() {
int ch;
while(1) {
printf("n1.Pushn 2.Popn 3.Displayn 4.Exit");
printf("nEnter your choice(1-4):"); scanf("%d",&ch);
switch(ch)
{
case 1: push(); break; case 2: pop(); break;
case 3: display(); break; case 4: exit(0);
default: printf("nWrong Choice!!"); return 0; } } }

126. void push()
{
int val;
if(top==MAX-1) printf("nStack is full!!");
else
{
printf("nEnter element to push:");
scanf("%d",&val);
top=top+1;
stack[top]=val;
}
}

127. Push operation in stack (Array)
n = 6
top = -1
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
Accept element 10
Increment top top = -1+1 = 0
top =0
Insert element in stack[top] 10

128. Push operation in stack (Array)
n = 6
top = 0
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
Accept element 20
Increment top top = 0+1 = 1 top =1
Insert element in stack[top] 10
20

129. Push operation in stack (Array)
n = 6
top = 1
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( ) top =2
10
20
30

130. Push operation in stack (Array)
n = 6
top = 2
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
top =3
10
20
30
40

131. Push operation in stack (Array)
n = 6
top = 3
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
top =4
10
20
30
40
50

132. Push operation in stack (Array)
n = 6
top = 4
stack[0]
stack
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
top =5
10
20
30
40
50
60

133. Push operation in stack (Array)
n = 6
top = 5
stack[0]
Stack is Full
stack[1]
stack[2]
stack[3]
stack[4]
stack[5]
Push ( )
top =5
10
20
30
40
50
60
if(top == n-1)
Print "Stack is Full"

134. void pop()
{
if(top==-1) printf("nStack is empty!!");
else
{
printf("nDeleted element is %d",stack[top]);
top=top-1;
}
}

135. void display()
{
int i;
if(top==-1) printf("nStack is empty!!");
else
{
printf("nStack is...n");
for(i=top; i >= 0 ;--i)
printf("%dn",stack[i]);
}
}

136. Display number of elements in stack
 static int count=0;
 Push function:
 After inserting an element, increment count
 Pop function
 After deleting an element, decrement count
 Display function
 Print “Count”

137. Stacks
 Demonstration of Stack using array

138. Stack using Linked List: Push Operation
no *next
node
Push ( )
10 NULL
(200)
(220)
20 (200)
(240)
30 (220)
(260)
30 (240)
top

139. Stack using Linked List: Pop Operation
10 NULL
(200)
(220)
20 (200)
(240)
30 (220)
(260)
30 (240)
top ptr

140. Stack using Linked List: Pop Operation
10 NULL
(200)
(220)
20 (200)
(240)
30 (220)
(260)
30 (240)
top
ptr

141. Stack using Linked List: Implementation
 Define node
struct Node
{
int data;
struct Node *next;
}*top = NULL;
void push(int);
void pop();
void display();

142. main() {
int choice, value;
while(1) {
printf("1. Pushn2. Popn3. Displayn4. Exitn");
printf("Enter your choice: "); scanf("%d",&choice);
switch(choice){
case 1: printf("Enter the value to be insert: ");
scanf("%d", &value); push(value);
break;
case 2: pop(); break;
case 3: display(); break;
case 4: exit(0);
default: printf("nWrong Option!!!n");
}
}
}

143. void push(int value)
{
struct Node *newNode;
newNode = (struct Node*)malloc(sizeof(struct Node));
newNode  data = value;
if(top == NULL)
newNodenext = NULL;
else
newNode  next = top;
top = newNode;
printf("nInsertion is Success!!!n");
}

144. void pop()
{
if(top == NULL)
printf("nStack is Empty!!!n");
else{
struct Node *temp = top;
printf("nDeleted element: %d", tempdata);
top = tempnext;
free(temp);
}
}

145. void display()
{
if(top == NULL) printf("nStack is Empty!!!n");
else{
struct Node *temp = top;
while(tempnext != NULL)
{
printf("%d--->",temp->data);
temp = temp  next;
}
printf("%d--->NULL",tempdata);
}
}

146. Stack using Linked List: Implementation
 Define node
typedef struct stack
{
int no;
struct stack *next;
} node;
node *top=NULL;
static int count=0;
void push();
void pop();
void display();

147. node *p;
p=(node *)malloc(sizeof(node));
printf("Enter no. to be pushed: "); scanf("%d",&p->no);
if(top==NULL) { pnext=NULL; top=p; count++; return; }
pnext=top;
top=p;
count++;
Push Operation: Implementation

148. Pop Operation: Implementation
node *ptr;
if(top==NULL) { printf("Stack is Empty"); return; }
ptr=top;
printf(“Popped number is: %d",ptrno);
top=topnext;
free(ptr);
count--;
}

149. Display Operation: Implementation
node *ptr;
printf("Total elements=%d",count);
if(top==NULL) { printf("nStack is empty"); return; }
ptr=top;
while(ptr != NULL)
{
printf("%d ",ptrno);
ptr=ptrnext;
}

150. Main function
void main()
{
int ch;
do
{
printf("n1.Push 2.Pop 3.Display 4.Exit");
printf("nEnter your choice: "); scanf("%d",&ch);
switch(ch) {case 1:push();break; case 2:pop(); break; case 3:display();
}
}
while(ch<4);
}

151. Unit 4
 Queue in Data structure
 Applications of Queues
 Operations on Queues
 Queue implementation using Arrays & LL
 Circular Queue

152. What is Queue
 Queue is a linear data structure in which the
insertion operation takes place at one end (Rear)
and deletion operation at the other end (Front).
 Queues can be implemented using either array or
linked list.
 Queue follows FIFO (First In First Out) principle

153. Queue Operations
 Enqueue
 Dequeue
 Display the elements of Queue
 Queues can be implemented using either Array of
Linked List

154. Queue using array (Insertion)
if (rear == MAX - 1)
printf("Queue is Full n");
else
{
if (front == - 1) front = 0;
printf("Insert element in queue : ");
scanf("%d", &value);
rear = rear + 1;
QueueA[rear] = value;
}
front = rear = -1
rear == MAX – 1
Print Queue Full front = -1 True: front = 0
Accept value
rear = rear + 1
QueueA[rear] = value
False:
True:

155. Queue using array (Deletion)
if (front == - 1 || front > rear)
{
printf("Queue is Empty n");
return ;
}
else
{
printf(“Deleted no. is : %dn",
QueueA[front]);
front = front + 1;
}
Print 'Deleted no. is', QueueA[front]
front = front + 1
Queue is empty when?
front = -1 front > rear
||
print Queue is Empty
False:
True:

156. Queue using Array (Display)
if (front == - 1)
printf("Queue is empty n");
else
{
printf("Queue is : n");
for (i = front; i <= rear; i++)
printf("%d ", QueueA[i]);
}
i = front i<=rear
Print QueueA[i]
i++
front = -1
Print "Queue is Empty"
for( ; ; )
False:
True:

157. Queue using Linked List: Insertion
10 NULL
no *next
node
(200)
front = NULL
rear = NULL
front = (200)
rear = NULL
(220)
20 NULL
10 (220)
front = (200)
rear = (220)
(240)
30 NULL
20 (240)
front = (200)
rear = (240)
(260)
40 NULL
30 (260)
front = (200)
rear = (260)

158. Queue using Linked List: Deletion
(200) (220)
20 NULL
(240)
30 NULL
20 (240)
(260)
40 NULL
30 (260)
front = (200)
rear = (260)
10 (220)
front = (200)
rear = (260)
ptr= front = (200)
10 (220)

159. Queue using Linked List: Deletion
(200) (220)
20 NULL
(240)
30 NULL
20 (240)
(260)
40 NULL
30 (260)
front = (220)
rear = (260)
10 (220)
front = (220)
rear = (260)
ptr = (200)
10 (220)

160. Queue using Linked List: Deletion
(200) (220)
20 NULL
(240)
30 NULL
20 (240)
(260)
40 NULL
30 (260)
front = (220)
rear = (260)
10 (220)
front = (220)
rear = (260)
ptr = (220)
10 (220)

161. Queue using Linked List: Deletion
(220)
20 NULL
(240)
30 NULL
20 (240)
(260)
40 NULL
30 (260)
front = (220)
rear = (260)
front = (220)
rear = (260)
ptr = (220)

162. Queue using Linked List: Deletion
(220)
20 NULL
(240)
30 NULL
20 (240)
(260)
40 NULL
30 (260)
front = (220)
rear = (260)
front = (240)
rear = (260)
ptr = (220)

163. Queue using Linked List: Deletion
(240)
30 NULL
(260)
40 NULL
30 (260)
front = (240)
rear = (260)
front = (240)
rear = (260)
ptr = (240)

164. Queue using Linked List
 Algorithm and Explanation for Insertion (Enqueue)
 Algorithm and Explanation for Deletion (Dequeue)
 Algorithm and Explanation for Display (Traversing)

165. Alg & Explanation for Insertion
no *next
node
Insertion()
1. Create node p and allocate memory
2. Accept no, store in p  no
3. Assign plink=NULL;
4. Check if rear=NULL
True:
a. front=rear=p;
b. return;
5. rear  link=p;
6. rear = p;
10 NULL
(200)
front = rear = NULL
front rear
(220)
20 NULL
10 (220)

166. Alg & Explanation for Insertion
no *next
node
Insertion()
1. Create node p and allocate memory
2. Accept no, store in p  no
3. Assign plink=NULL;
4. Check if rear=NULL
True:
a. front=rear=p;
b. return;
5. rear  link=p;
6. rear = p;
10 NULL
(200)
front = rear = NULL
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)

167. Alg & Explanation for Insertion
no *next
node
Insertion()
1. Create node p and allocate memory
2. Accept no, store in p  no
3. Assign plink=NULL;
4. Check if rear=NULL
True:
a. front=rear=p;
b. return;
5. rear  link=p;
6. rear = p;
10 NULL
(200)
front = rear = NULL
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)

168. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
10 NULL
(200)
front = (200)
rear = (240)
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)

169. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
10 NULL
(200)
front = (200)
rear = (240)
ptr = (200)
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)
ptr

170. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
10 NULL
(200)
front = (220)
rear = (240)
ptr = 200
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)
ptr

171. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
10 NULL
(200)
front = (220)
rear = (240)
front rear
(220)
20 NULL
10 (220)
(240)
30 NULL
20 (240)
ptr

172. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
front = (220)
rear = (240)
ptr = (220)
front rear
(220)
20 NULL
(240)
30 NULL
20 (240)
ptr

173. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
front = (240)
rear = (240)
ptr = (220)
front
rear
(220)
20 NULL
(240)
30 NULL
20 (240)
ptr

174. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
front = (240)
rear = (240)
ptr = (240)
front
rear
(240)
30 NULL
ptr

175. Alg & Explanation for Deletion
no *next
node
Deletion
1. if(front==NULL)
a. Print “Queue Empty”
b. Return
2. ptr = front;
3. front=frontlink;
4. if(front==NULL)
a. rear=NULL;
5. Print ptrno;
6. free(ptr);
front = NULL
rear = NULL
ptr = (240)
front
rear
(240)
30 NULL
ptr

176. Algorithm for Display
if(front == NULL)
{
Print “Queue is empty”;
return;
}
ptr = front;
while(ptr != NULL) { Print ptrno; ptr = ptrlink; }
}

177. Queue using Linked List: Implementation
 Define node
typedef struct queue
{
int no;
struct queue *link;
}node;
void insertion();
void deletion();
void display();
node *front=NULL,*rear=NULL;

178. insertion ( )
void insertion()
{
node *p; p=malloc(sizeof(node));
// Accept no and store in pno);
plink=NULL;
if(rear==NULL) { front=rear=p; return; }
rearlink=p;
rear=p;
}

179. Deletion ( )
void deletion()
{
node *ptr;
if(front == NULL) { printf("nUnderflow"); return; }
ptr = front; front = frontlink;
if(front == NULL) rear = NULL;
printf("nDeleted number is: %d",ptr  no);
free(ptr);
}

180. Display ( )
void display()
{
node *ptr;
if(front == NULL) { Print “Queue is empty”; return; }
ptr = front;
while(ptr != NULL) { Print ptrno; ptr = ptrlink;
}
}

181. Main function
void main()
{
int ch;
do
{
printf("n1.Insert 2.Delete 3.Display 4.Exit");
printf("nEnter your choice: "); scanf("%d",&ch);
switch(ch) { case 1:insertion();break; case 2:deletion();break;
case 3:display(); } }
while(ch<4);
}

182. Unit 5: Non Linear Data Structures
 Tree: Basic terminology, Binary Trees
 Representation of Binary Tree using array and LL
 Binary Search Trees
 Operations on BST: Insertion, Deletion
 Recursive Traversals: PreOrder, InOrder, PostOrder

183. Representation of binary tree using array

184. Sequential representation of Binary Tree
 For first case(0 to n-1),
if root is indicated by „p‟;
then left child=(2*p)+1;
and right child=(2*p)+2;
 For second case(1to n),
if root=p;
then left child=(2*p);
and right child=(2*p)+1;
where root, left child and
right child are the values of
indices of the array.

185. Representation of
binary tree using array

186. Representation
of
binary tree
using array

188. Sequential representation
 Advantages:
 best representation for complete and full binary tree
representation
 Disadvantages:
 Height of tree should be known
 Memory may be wasted
 Insertion and deletion of a node is difficult

189. Binary Tree:
Linked List
Representation
left right
Root
left right
data
left right
data
left right
data
left right
data
left right
data
left right
data

190. struct node
{
int data;
struct node *left;
struct node *right;
};
/* Initialize nodes */
struct node *root;
struct node *one = NULL;
struct node *two = NULL;
struct node *three = NULL;
Binary Tree: Linked List Representation

191. Binary Tree: Linked List Representation
/* Allocate memory */
one = malloc(sizeof(struct node));
two = malloc(sizeof(struct node));
three = malloc(sizeof(struct node));
/* Assign data values */
Onedata = 1;
two  data = 2;
three  data = 3;
/* Connect nodes */
oneleft = two; one  right = three;
two  left = NULL;
two  right = NULL;
three  left = NULL;
three  right = NULL;
/* Save address of first node in root */
root = one;

192. Binary Tree: Linked List Representation
 Advantages:
 Height of tree need not be known
 No memory wastage
 Insertion and deletion of a node is done without affecting other
nodes
 Disadvantages:
 Direct access to node is difficult
 Additional memory required for storing address of left and right
node

193. Binary Search Tree
 A binary tree T is termed as binary search tree (or
binary sorted tree) if each node N of T satisfies the
following property:
The value at N is a greater than every value in the
left sub-tree of N and is less than every value in the
right sub-tree of N.

194. Binary Search Trees (BSTs)
 Binary Search Tree Property:
The value stored at
a node is greater than
the value stored at its
left child and less than
the value stored at its
right child

195. Construct BST: 10, 12, 5, 4, 20, 8, 7, 15, 13

196. BST Example
 Construct the Binary Search Tree for the given
elements:
 50, 17, 72, 12, 23, 54, 76, 9, 14, 19, 67

197. Search an element in BST
 Algorithm:
1. If (root == NULL) return NULL;
2. If (num == rootdata) return root  data;
3. If (num < root  data)
return search(root  left)
4. If (num > root  data )
return search(root  right)

198. struct node
{
int data;
struct node* left;
struct node* right;
};
struct node* insert(struct node* root, int data)
{
if (root == NULL) return createNode(data);
if (data < root  data)
root  left = insert(root  left, data);
else if (data > root  data)
root  right = insert(root  right, data);
return root;
}
struct node* createNode(value) {
struct node* newNode=malloc(sizeof(struct node));
newNodedata = value;
newNode  left = NULL;
newNode  right = NULL;
return newNode; }
int main()
{
struct node *root = NULL;
root = insert(root, 8); insert(root, 3);
insert(root, 1); insert(root, 6);
insert(root, 7); insert(root, 10);
insert(root, 14); insert(root, 4); }

199. Insert value in Binary Search Tree(BST)
 Algorithm
1. If (node == NULL) return createNode(data)
2. if (data < nodedata)
node  left = insert(node  left, data);
3. else if (data > node  data)
node  right = insert(node  right, data);
return node;

200. Find the largest element in BST
if ( root == null) { return NULLL; }
struct node *currNode = root;
while(currNoderight != null)
{
currNode = currNoderight;
}
return currNodedata;

201. Deleting a node in BST
 Three cases in deleting a node:
 Case 1: Deleting a Leaf node (A node with no children)
 Case 2: Deleting a node with one child
 Case 3: Deleting a node with two children

202. Case 1: Deleting a leaf node
 Step 1 - Find the node to be deleted using search operation
 Step 2 - Delete the node using free function (If it is a leaf) and
assign NULL to the Parent‟s left or right accordingly.

203. Case 2: Deleting a node with one child
 Step 1 - Find the node to be deleted using search
operation
 Step 2 - If it has only one child then create a link
between its parent node and child node.
 Step 3 - Delete the node using free function and
terminate the function.

204. Deleting a node with one child

205. Find the smallest element in BST
if ( root == null ) return NULL;
struct node *currNode = root;
while(currNodeleft != null)
{
currNode = currNodeleft;
}
return currNodedata;

206. Case 3: Deleting a node with two children
Step1: Find the deletion node p (the node that we want to delete)
Step 2: Find the successor node of p
2.1 Successor node is the node in the right subtree that has the
minimum value (Or)
2.1 Successor node is the node in the left subtree that has the
maximum value
Step 3: Replace the content of node p with the content of the successor
node
Step 4: Delete the successor node

207. Deleting a node with 2 children

208. Recursive Tree Traversals
 InOrder
 PreOrder
 PostOrder

209. 209
Preoder, Inorder, Postorder
 In Preorder, the root
is visited before (pre)
the subtrees traversals
 In Inorder, the root is
visited in-between left
and right subtree traversal
 In Postorder, the root
is visited after (post)
Preorder Traversal:
1. Visit the root
2. Traverse left subtree
3. Traverse right subtree
Inorder Traversal:
1. Traverse left subtree
2. Visit the root
3. Traverse right subtree
Postorder Traversal:
1. Traverse left subtree
2. Traverse right subtree
3. Visit the root

210. InOrder Tree Traversal
void InOrder (node *ptr)
{
if(ptr == NULL) return;
else
{
InOrder(ptrleft);
Print ptr  value
InOrder (ptr  right);
}
}

211. InOrder: Example 5

212. Pre Order Traversal
void PreOrder(node *ptr)
{
if(ptr == NULL) return;
else
{
Print ptr  value;
PreOrder (ptr  left);
PreOrder (ptr  right);
}
}