Trees
Question 1 |
The postorder traversal of a binary tree is 8, 9, 6, 7, 4, 5, 2, 3, 1. The inorder traversal of the same tree is 8, 6, 9, 4, 7, 2, 5, 1, 3. The height of a tree is the length of the longest path from the root to any leaf. The height of the binary tree above is _______.
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
Post – 8 9 6 7 4 5 2 3 1
In – 8 6 9 4 7 2 5 1 3
Post: 8 9 6 7 4 5 2 3 1→(root)
In: 8 6 9 4 7 2 5→(left subtree) 13→(right subtree)

In – 8 6 9 4 7 2 5 1 3
Post: 8 9 6 7 4 5 2 3 1→(root)
In: 8 6 9 4 7 2 5→(left subtree) 13→(right subtree)




Question 2 |
Let T be a tree with 10 vertices. The sum of the degrees of all the vertices in T is _________.
18 | |
19 | |
20 | |
21 |
Question 2 Explanation:
Sum of degrees of all vertices is double the number of edges.
A tree, with 10 vertices, consists n - 1, i.e. 10 - 1 = 9 edges.
Sum of degrees of all vertices = 2(#edges)
= 2(9)
= 18
A tree, with 10 vertices, consists n - 1, i.e. 10 - 1 = 9 edges.
Sum of degrees of all vertices = 2(#edges)
= 2(9)
= 18
Question 3 |
The number of ways in which the numbers 1, 2, 3, 4, 5, 6, 7 can be inserted in an empty binary search tree, such that the resulting tree has height 6, is _________.
Note: The height of a tree with a single node is 0.64 | |
65 | |
66 | |
67 |
Question 3 Explanation:
To get the tree of height 6, every level should contain only 1 node.
So to get such tree at each level we should have either maximum or minimum element from remaining numbers till that level. So no. of binary search tree possible is,
1st level - 2 (maximum or minimum)
⇓
2nd level - 2
⇓
3rd level - 2
⇓
4th level - 2
⇓
5th level - 2
⇓
6th level - 2
⇓
7th level - 2
= 2 × 2 × 2 × 2 × 2 × 2 × 1
= 26
= 64
So to get such tree at each level we should have either maximum or minimum element from remaining numbers till that level. So no. of binary search tree possible is,
1st level - 2 (maximum or minimum)
⇓
2nd level - 2
⇓
3rd level - 2
⇓
4th level - 2
⇓
5th level - 2
⇓
6th level - 2
⇓
7th level - 2
= 2 × 2 × 2 × 2 × 2 × 2 × 1
= 26
= 64
Question 4 |
The height of a tree is the length of the longest root-to-leaf path in it. The maximum and minimum number of nodes in a binary tree of height 5 are
63 and 6, respectively | |
64 and 5, respectively | |
32 and 6, respectively
| |
31 and 5, respectively |
Question 4 Explanation:
Maximum number of nodes in a binary tree of height h is,
2h+1 - 1 = 25+1 - 1 = 63
Minimum number of nodes in a binary tree of height h is
h + 1 = 5 + 1 = 6
2h+1 - 1 = 25+1 - 1 = 63
Minimum number of nodes in a binary tree of height h is
h + 1 = 5 + 1 = 6
Question 5 |
Consider the following 2-3-4 tree (i.e., B-tree with a minimum degree of two) in which each data item is a letter. The usual alphabetical ordering of letters is used in constructing the tree.
What is the result of inserting G in the above tree ?

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None of the above |
Question 5 Explanation:
Given Tree is

Insert G at root level:

Insert G at root level:

Question 6 |
A binary tree T has n leaf nodes. The number of nodes of degree 2 in T is:
log2 n | |
n - 1 | |
n | |
2n |
Question 6 Explanation:
A binary tree is a tree data structure in which each node has atmost two child nodes.
The no. of subtrees of a node is called the degree of the node. In a binary tree, all nodes have degree 0, 1 and 2.
The degree of a tree is the maximum degree of a node in the tree. A binary tree is of degree 2.
The number of nodes of degree 2 in T is "n - 1".
The no. of subtrees of a node is called the degree of the node. In a binary tree, all nodes have degree 0, 1 and 2.
The degree of a tree is the maximum degree of a node in the tree. A binary tree is of degree 2.
The number of nodes of degree 2 in T is "n - 1".
Question 7 |
4 | |
5 | |
6 | |
7 | |
Both A and D |
Question 7 Explanation:
Case 1:

Where L is leaf node.
So, no. of internal node is 4.
Case 2:

Where L is leaf node.
So, no. of internal node is 7.

Where L is leaf node.
So, no. of internal node is 4.
Case 2:

Where L is leaf node.
So, no. of internal node is 7.
There are 7 questions to complete.