Height of Node MCQ Quiz - Objective Question with Answer for Height of Node - Download Free PDF

The correct answer is option 2: 2^k+1 − 1

Key Points

• In a binary tree, each level can contain at most twice the number of nodes as the previous level.
• If height k means the root is at level 0 and the last level is k, then total levels = k + 1.
• The maximum number of nodes in such a binary tree is given by the sum of the geometric progression:
  • Nodes = 1+2+4+⋯+2k=2k+1−1" id="MathJax-Element-203-Frame" role="presentation" style="position: relative;" tabindex="0">1+2+4+⋯+2k=2k+1−11+2+4+⋯+2k=2k+1−1 $1 + 2 + 4 + \dots + 2^k = 2^{k+1} - 1$
• This is the formula for the number of nodes in a perfect binary tree of height k.

Additional Information

• 2^k − 1: This would be valid if height was defined as number of levels minus one.
• 2^k−1 + 1 and 2^k + 1: Incorrect and do not follow from the geometric progression formula.
• Therefore, for height k starting from root at 0, maximum nodes = 2k+1−1" id="MathJax-Element-204-Frame" role="presentation" style="position: relative;" tabindex="0">2k+1−12k+1−1 $2^{k+1} - 1$

Hence, the correct answer is: option 2: 2^k+1 − 1

Concepts:

Minimum height of the tree is when all the levels of BST are completely filled.

Maximum height of the binary search tree (BST) is the worst case when nodes are in skewed manner.

Formula:

Minimum height of the BST with n nodes is ⌈log₂ (n + 1)⌉ - 1

Maximum height of the BST with n nodes is n - 1.

Calculation:

Maximum height of the BST with 15 nodes 15 - 1 = 14

Diagram:

Minimum height of the BST with n nodes is ⌈log₂ (15 + 1)⌉ - 1 = 3

Diagram: 

Graph with height 5:

The maximum number of nodes present: 63 

Tips and Tricks:

If n is number of nodes and h is minimum height of in a binary search tree, then

n = 2^h+1 – 1

n = 2⁵⁺¹ – 1

Concepts:

Minimum height of the tree is when all the levels of BST are completely filled.

Maximum height of the binary search tree (BST) is the worst case when nodes are in skewed manner.

Formula:

Minimum height of the BST with n nodes is ⌈log₂ (n + 1)⌉ - 1

Maximum height of the BST with n nodes is n - 1.

Calculation:

Maximum height of the BST with 15 nodes 15 - 1 = 14

Diagram:

Minimum height of the BST with n nodes is ⌈log₂ (15 + 1)⌉ - 1 = 3

Diagram: 

Concepts:

Minimum height of the tree is when all the levels of BST are completely filled.

Maximum height of the binary search tree (BST) is the worst case when nodes are in skewed manner.

Formula:

Minimum height of the BST with n nodes is ⌈log₂ (n + 1)⌉ - 1

Maximum height of the BST with n nodes is n - 1.

Calculation:

Maximum height of the BST with 15 nodes 15 - 1 = 14

Diagram:

Minimum height of the BST with n nodes is ⌈log₂ (15 + 1)⌉ - 1 = 3

Diagram: 

The correct answer is option 2: 2^k+1 − 1

Key Points

• In a binary tree, each level can contain at most twice the number of nodes as the previous level.
• If height k means the root is at level 0 and the last level is k, then total levels = k + 1.
• The maximum number of nodes in such a binary tree is given by the sum of the geometric progression:
  • Nodes = 1+2+4+⋯+2k=2k+1−1" id="MathJax-Element-203-Frame" role="presentation" style="position: relative;" tabindex="0">1+2+4+⋯+2k=2k+1−11+2+4+⋯+2k=2k+1−1 $1 + 2 + 4 + \dots + 2^k = 2^{k+1} - 1$
• This is the formula for the number of nodes in a perfect binary tree of height k.

Additional Information

• 2^k − 1: This would be valid if height was defined as number of levels minus one.
• 2^k−1 + 1 and 2^k + 1: Incorrect and do not follow from the geometric progression formula.
• Therefore, for height k starting from root at 0, maximum nodes = 2k+1−1" id="MathJax-Element-204-Frame" role="presentation" style="position: relative;" tabindex="0">2k+1−12k+1−1 $2^{k+1} - 1$

Hence, the correct answer is: option 2: 2^k+1 − 1

Graph with height 5:

The maximum number of nodes present: 63 

Tips and Tricks:

If n is number of nodes and h is minimum height of in a binary search tree, then

n = 2^h+1 – 1

n = 2⁵⁺¹ – 1

Insert B - 2 split required

Insert I - 1 split required

Insert K – 0 split required