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