If the number of leaves in a strictly binary tree is an odd number, then what…

Concept

A strictly binary tree (a full binary tree) is one in which every node has exactly 0 or 2 children. If L is the number of leaf nodes (0 children) and I is the number of internal nodes (exactly 2 children), the identity L = I + 1 always holds for such trees, and the total node count is N = L + I.

Application

1. From L = I + 1, we get I = L − 1.

2. Total nodes: N = L + I = L + (L − 1) = 2L − 1.

3. The stem gives L as odd. Substituting into N = 2L − 1: 2L is always even (2 times any integer), so N = (even number) − 1 is always odd.

4. So whenever L is odd, N = 2L − 1 is guaranteed to be odd.

Cross-check

Take a concrete strictly binary tree with L = 3 leaves (odd). Then I = L − 1 = 2 internal nodes, so N = 5, an odd number — matching the derivation. A minimal case also checks out: a single-node tree has L = 1 leaf (odd), I = 0, N = 1, also odd.

This also rules out the other statements: since N = 2L − 1 is strictly less than 2L, N can never exceed twice the leaf count. Since N = 1 = L when L = 1, N can equal the leaf count in at least one valid strictly binary tree. And since N = 2L − 1 is always odd, N is never even.

Result

The total number of nodes in the tree is always an odd number.