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 O (i) O / O / O (ii) O / O \ O (iii) O \ O \ O (iv) O \ O / O (v)

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: http://en.wikipedia.org/wiki/Reverse_Polish_notation#The_postfix_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

1

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. a / \ / \ b c / \ / \ / \ / \ d e f g

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