Data structures - ISC Class 12 Computer Science Questions with Answers, Page 5

List the leaf nodes of the tree.

Write the postorder traversal of the above tree structure.

State the level numbers of the nodes N and R if the root is at 0 (zero) level.

Queue is an entity which can hold a maximum of 100 integers. The queue enables the user to add integers from the rear and remove integers from the front. Define a class Queue with the following details: Class name : Queue Data Members / instance variables: Que[ ] : array to hold the integer elements size : stores the size of the array front : to point the index of the front rear : to point the index of the rear Member functions: Queue (int mm) : constructor to initialize the data size = mm, front = 0, rear = 0 void addele(int v ) : to add integer from the rear if possible else display the message “Overflow” int delele( ) : returns elements from front if present, otherwise displays the message “Underflow” and return -9999 void display ( ) : displays the array elements Specify the class Queue giving details of ONLY the functions void addele(int) and int delele( ). Assume that the other functions have been defined. The main function and algorithm need NOT be written.

void addele(int v)
{
    if (rear == size)
        System.out.println("Overflow");
    else
    {
        Que[rear] = v;
        rear++;
    }
}

int delele()
{
    if (front == rear)
    {
        System.out.println("Underflow");
        return -9999;
    }
    else
    {
        int v = Que[front];
        front++;
        return v;
    }
}

Convert the following infix expression to postfix form: $P * Q / R + (S + T)$

List the internal nodes of the right sub-tree.

A linked list is formed from the objects of the class Node. The class structure of the Node is given below:

class Node
{
    int num;
    Node next;
}

Write an Algorithm OR a Method to count the nodes that contain only odd integers from an existing linked list and returns the count. The method declaration is as follows:

int CountOdd( Node startPtr )

int CountOdd(Node startPtr)
{
    int count = 0;
    Node ptr = startPtr;
    while (ptr != null)
    {
        if (ptr.num % 2 != 0)
            count++;
        ptr = ptr.next;
    }
    return count;
}