Skip to content
Related Articles

Related Articles

Data Structures and Algorithms | Set 10
  • Difficulty Level : Basic
  • Last Updated : 27 Mar, 2017

Following questions have been asked in GATE CS 2007 exam.

1. The height of a binary tree is the maximum number of edges in any root to leaf path. The maximum number of nodes in a binary tree of height h is:
(A) 2^h -1
(B) 2^(h-1) – 1
(C) 2^(h+1) -1
(D) 2*(h+1)

Answer (C)
Maximum number of nodes will be there for a complete tree.
Number of nodes in a complete tree of height h = 1 + 2 + 2^2 + 2*3 + …. 2^h = 2^(h+1) – 1

2: The maximum number of binary trees that can be formed with three unlabeled nodes is:
(A) 1
(B) 5
(C) 4
(D) 3

Answer (B)

          /     \
        O        O





Note that nodes are unlabeled. If the nodes are labeled, we get more number of trees.

3. Which of the following sorting algorithms has the lowest worst-case complexity?
(A) Merge sort
(B) Bubble sort
(C) Quick sort
(D) Selection sort

Answer (A)

Worst case complexities for the above sorting algorithms are as follows:
Merge Sort — nLogn
Bubble Sort — n^2
Quick Sort — n^2
Selection Sort — n^2

4. The following postfix expression with single digit operands is evaluated using a stack:

              8 2 3 ^ / 2 3 * + 5 1 * - 

Note that ^ is the exponentiation operator. The top two elements of the stack after the first * is evaluated are:
(A) 6, 1
(B) 5, 7
(C) 3, 2
(D) 1, 5

Answer (A)
The algorithm for evaluating any postfix expression is fairly straightforward:

1. While there are input tokens left
    o Read the next token from input.
    o If the token is a value
       + Push it onto the stack.
    o Otherwise, the token is an operator 
      (operator here includes both operators, and functions).
       * It is known a priori that the operator takes n arguments.
       * If there are fewer than n values on the stack
        (Error) The user has not input sufficient values in the expression.
       * Else, Pop the top n values from the stack.
       * Evaluate the operator, with the values as arguments.
       * Push the returned results, if any, back onto the stack.
2. If there is only one value in the stack
    o That value is the result of the calculation.
3. If there are more values in the stack
    o (Error)  The user input has too many values.

Source for algorithm:

Let us run the above algorithm for the given expression.
First three tokens are values, so they are simply pushed. After pushing 8, 2 and 3, the stack is as follows

    8, 2, 3  

When ^ is read, top two are popped and power(2^3) is calculated

    8, 8 

When / is read, top two are popped and division(8/8) is performed


Next two tokens are values, so they are simply pushed. After pushing 2 and 3, the stack is as follows

    1, 2, 3

When * comes, top two are popped and multiplication is performed.

    1, 6

5. The inorder and preorder traversal of a binary tree are d b e a f c g and a b d e c f g, respectively. The postorder traversal of the binary tree is:

(A) d e b f g c a
(B) e d b g f c a
(C) e d b f g c a
(D) d e f g b c a

Answer (A)

Below is the given tree.
                           /    \
                        /          \
                      b             c
                   /   \          /   \
                 /       \      /       \
               d         e    f          g

Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc.

Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :