mid-term exam - sample solution

8/13/2019 Mid-Term Exam - Sample Solution

1/6

UNIVERSITY OF THE WEST INDIES

CAVE HILL CAMPUS

Department of Computer Science, Mathematics & Physics

COMP2115Information Structures

Mid-Term Examination

SAMPLE SOLUTION

12 November, 2010 Time: 1 hrs

INSTRUCTIONS: This paper contains TWO (2) pages and FOUR (4) questions

DO ALL FOUR QUESTIONS

Question 1

(a) Define an Abstract Data Type (ADT). [2 marks]

A data structure and a collection of functions or procedures which operate on the

data structure.

(b) Identify the fundamental characteristic(s) which an ADT must possess. [2 marks]

1. Be able to create a new collection

2. Be able add an item to a collection

3. Be able add to delete an item from a collection4. Be able add to find an item matching some criterion in the collection5. Be able add to destroy the collection

(c) How does an ADT differ from other data containers such as arrays? [1 mark]

Non-ADT data containers such as arrays, are static structure which require

contiguous memory locations to be allocated to them as soon as they are declared.

The size of such a structure does not change throughout the programs execution.The ADT data containers are dynamic data structures and are the exact opposite.

Memory locations allocated to ADTs need not be contiguous and the structure is

allowed to expand and shrink as the data needs change.

(d) Identify the difference(s) between a Constructorfunction and a Destructorfunction in a

class definition. [2 marks]

Constructor : Initialize objects of a class during instantiation.

Destructor : Performs termination house-keeping just before the object is destroyed

or goes out of scope.

(e) Briefly describe accessor, mutatorand utilityfunctions in classes. [3 marks]Accessor: Member functions which access or get the data from within an objects

attribute, but do not change the data.

Mutator: Member functions which access the attributes within an in order to

change the data value.

Utility : Private class member fuctions which are called by other member

functions to perform a particular task within the object.


2/6

Question 2

(a) Briefly differentiate between the following Abstract Data Types:

(i) Queue ANDPriority Queue [2 marks]

Queue:

Lists which employ a first-in, first-out (FIFO) concept. Nodes are removed

only from the head and nodes are inserted only at the tail.

Priority Queue:

A lists which imposes a sorted priority on the elements so that the elementwith the highest priority is always at the head of the list. This sorting is done

whenever a new element is added to the list. Nodes are removed only from the

headi.e. the node with the highest priority.

(ii) Stack ANDDeque [2 marks]

Stack:

Lists which employ a last-in, first-out (LIFO) concept. Nodes are added

and removed only from the top of the structure.

Deque:

A double ended queue which permits insertion and deletion at both the

head and tail of the list. Insertion and deletion are only permitted at thetwo ends.

(iii) Binary Search Tree ANDBinary Heap [2 marks]

Binary Search Tree:

A binary tree such that all elements stored in the left subtree of a node

whose value is K have values less than K. All elements stored in the

right subtree of a node whose value is K have values greater than or

equal to K. All sub-trees are BSTs.

Binary Heap:

A special kind of binary tree which has two properties that are not

generally true for other trees:

1. Completeness

The tree is complete, which means that nodes are added from

top to bottom, left to right, without leaving any spaces. A binary

tree is completely full if it is of height, h, and has 2h+1-1 nodes.

2. Heapness

The item in the tree with the highest priority is at the top of the

tree, and the same is true for every subtree.

(b) With reference to a Tree ADT, identify the difference(s) between the following pairs of

terms:(i) Heightand Depth [2 marks]

The depth of a node M in a tree is the length of the path from the root ofthe tree to M.

The height of a tree is one more than the depth of the deepest node in thetree.


3/6

(ii) Leaf Nodeand Internal Node [2 marks]

A node, apart from the root) which has no children is called a leaf node orsimply a leaf.

A node (apart from the root) that has children is an internal node.(iii) Visitingand Traversing [2 marks]

A node is visited when program control arrives at the node, usually for thepurpose of carrying out some operation on the node.

To traverse a tree means to visit all of the nodes in some specified order.(iv) Completebinary trees and Fullbinary trees [2 marks]

A full binary tree is a binary tree where each node is either a leaf or is aninternal node with exactly two non-empty children.

A complete binary tree is a binary tree where if the height is d, and alllevels, except possibly level d, are completely full. If the bottom level is

incomplete, then it has all nodes to the left side.

(c) Describe balance with respect to trees. [1 mark]

Balance is described as the comparison between the heights of a nodes left and right

sub-trees. If the heights are equal, then the node is said to be balance. Otherwise, it

may be said to be left-heavy or right-heavy.

Question 3

Consider the following data set: 53, 24, 30, 62, 28, 60, 23, 61, 70, 2, 7, 31, 44, 3, 16.

(a) Insert the values into a Binary Search Tree. [2 marks]

(b) From the tree structure created in (a) above, give the results of:

(i) Inordertraversal 2, 3, 7, 16, 23, 24, 28, 30, 31, 44, 53, 60, 61, 62, 70

(ii) Preordertraversal 53, 24, 23, 2, 7, 3, 16, 30, 28, 31, 44, 62, 60, 61, 70(iii) Postordertraversal 3, 16, 7, 2, 23, 28, 44, 31, 30, 24, 61, 60, 70, 62, 53

[3 marks]

53

62

7060

24

30

28 31

23

2

7

3 16

44

61


4/6


5/6

(a) addNode [6 marks]

void PriorityQueue::addNode(int val)

{

Node *newnode= new Node(val);

if (Head == NULL) Head = newnode;

else

{if ( val > Head->getData())

{

Newnode->setNext(Head);

Head = newnode;

}

else

{

Node *prevPtr = Head;

Node *curPtr = Head->getNext();

while(curPtr!=NULL && curPtr->getData()> val){

prevPtr = curPtr;

curPtr = curPtr->getNext();

}

if ( curPtr == NULL) precPtr->setNext(newnode);

else

{

Newnode->setNext(curPtr);

prevPtr->setNext(newnode);

}}

}

(b) dequeue [4 marks]

int PriorityQueue::dequeue()

{

Node* temp = Head;

int val;

if (Head == NULL) return -1;

else{

Temp = Head;

Head = Head->getNext();

Val = temp->getData();

delete temp;

return val;

}

}


6/6

(c) find [4 marks]

bool PriorityQueue::find(int val)

{

if (Head == NULL)

return False;else

{

Node* temp;

for (temp = Head; temp != NULL && temp->getData() != val; temp =

temp->getNext();

}

if (temp == NULL)

return False;

else

return True;

}

(d) isEmpty [1 mark]

bool PriorityQueue::isEmpty()

{

if (Head == NULL)

return true;

else

return false;

}

Note:These are class member functions and NOT regular functions.

mid-term exam - sample solution

Documents