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CUSTOMER_CODE
SMUDE
DIVISION_CODE
SMUDE
EVENT_CODE
Jan2017
ASSESSMENT_CODE MIT102_Jan2017
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
12138
QUESTION_TEXT
What is data structure? Discuss briefly on types of data structures.
SCHEME OF
EVALUATION
The data structure represents the logical relationship of the particular
data sets (1 mark)
2 types
1. linear data structures
2. Non linear data structures (2 marks)
Linear data structure
When the data is stored in the memory in linear or sequential form is
called linear data structure (1 mark)
Examples of linear data structures include. Array, stack, queue,
linked list
Array Expn
Stack Expn
Queue Expn
Linked list Expn (any 3=3 marks)
Non linear data structure:
When the data is stored in the memory in dispersed or non sequential
order is called non linear data structure
(1 mark)
Example:
Tress Expn (1 mark)
Graphs Expn (1 mark)
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73331
QUESTION_TEXT
Give an account of types of mass storage devices
SCHEME OF
EVALUATION
Floppy disks: Relatively slow and have a small capacity, but they are
portable, inexpensive, and universal.
* Hard disks: Very fast and with more capacity than floppy disks, but
also more expensive. Some hard disk systems are portable (removable
cartridges), but most are not.
* Optical disks: Unlike floppy and hard disks, which use
electromagnetism to encode data, optical disk systems use a laser to
read and write data. Optical disks have very large storage capacity, but
they are not as fast as hard disks. In addition, the inexpensive optical
disk drives are read-only. Read/write varieties are expensive.
* Tapes: Relatively inexpensive and can have very large storage
capacities, but they do not permit random access of data.
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73333
QUESTION_TEXT
Explain the array implementation of stack.
SCHEME OF
EVALUATION
Push operation:
5 Marks
PUSH (STACK, TOP, MAXSTK, ITEM)
1. [STACK ALREADY FULL?]
If TOP=MAXSTK, THEN PRINT ‘OVERFLOW’ AND
RETURN;
2. SET TOP=TOP+1
3. SET STACK[TOP]=ITEM
4. RETURN
Pop operation:
5 Marks
POP (STACK, TOP, ITEM)
1. [STACK IS EMPTY?]
IF TOP=-1, THEN PRINT ‘STACK EMPTY’ AND
RETURN
2. SET ITEM = STACK[TOP]
3. SET TOP=TOP-1
4. RETURN
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
73334
QUESTION_TEXT
Give the recursive functions for the binary tree traversals.
A binary tree is a special case of tree where no node of a tree can have
a degree of more than two. Therefore, a binary tree is a set of zero or
more nodes such that:
SCHEME OF
EVALUATION
There is a specially designated node called the root of the tree
The remaining nodes are partitioned into two disjointed sets, T1 and
T2, each of which is a binary tree. T1 is called the left sub tree and T2
is called right sub tree.
Binary tree traversals:
Inorder traversal:
Void print_inorder(tree_node *p)
{
If (p!=NULL)
{
Print_inorder(pleft);
Cout< < p- >data< < endl;
Print_inorder(pright);
}
3 Marks
}
Postorder traversal:
Void print_postorder(tree_node *p)
{
If (p!=NULL)
{
Print_postorder(pleft);
Print_postorder(pright);
Cout< < p- >data< < endl;
}
3.5 Marks
}
Preorder traversal:
Void print_preorder(tree_node *p)
{
If (p!=NULL)
{
Cout< < p- >data< < endl;
Print_preorder(pleft);
Print_preorder(pright);
}
3.5 Marks
}
QUESTION_TYPE
DESCRIPTIVE_QUESTION
QUESTION_ID
120229
QUESTION_TEXT
Write an algorithm to insert an element into a queue using arrays
and explain.
QINSERT(QUEUE,N,FRONT,REAR,ITEM)
1. If FRONT=1 and REAR=N or FRONT=REAR+1 then
Write OVERFLOW and return
2. [Find new value of REAR]
If FRONT:=NULL then
Set FRONT:=1 and REAR:=1
Else if REAR=N then
Else if REAR=N then
Set REAR:=1
Else
Set REAR:=REAR+1
3. Set QUEUE[REAR]:=ITEM
SCHEME OF
EVALUATION
4.
Return
(5 marks)
Explanation (5 marks)
QUESTION_TYP
DESCRIPTIVE_QUESTION
E
QUESTION_ID
120234
QUESTION_TEX
Write the steps for converting the general tree to a binary tree.
T
1. The root of the general tree must be the root of the binary tree
(
SCHEME OF
EVALUATION
1 mark)
2. Determine the first child of the root which is the left most node in the general
tree at the next level.
(1 mark)
3. Insert this node.
The child-parent relationship in general tree would be considered in binary tree
also
(1 mark)
4. Continue finding the first child of each parent node and insert it below the
parent child
(1 mark)
5. When no more first children exist in the path just used, move back to the
parent of the last node entered and determine the next sibling, then insert right
to the previous left sibling (where sibling in general tree would be child in binary
tree)
(
3 marks)
6. In order to complete the tree, the following steps to be executed:
→Look for the completion of sibling at one generation level
→Next move to left most child of the successor generation and repeat
the process as same until the sibling of the generation are
inserted
(3 marks)