Question

Prove that the number of nodes in a binary decision tree will be full with k levels if and only if the number of nodes available is 2^k - 1.

Note that to conduct this proof, you will need to prove the statement both ways.

Answer 1

As we know, the maximum number of a child of a tree can have is either 2 or 0.

The maximum number of nodes in a binary Decision tree or we can say (BST) of depth K is 2 ^ K -1, where, K >=1. Consider only this case.

This can easily be proved by PMI (Principle Mathematical Induction).

Suppose, we are at level 1. Now put the value of K = 1 in Equation (2 ^ K- 1), the answer is coming out be 1 for Level 1. At level 1 no further child is present.

Suppose, we are at level 3, Number of possible child Level 3 can have is (2 ^ 3 - 1 ) which is nothing but 7.