What is the maximum number of nodes at level L in a binary tree, assuming the…

Concept: In a binary tree every node has at most two children. If levels are numbered starting from 0 at the root, a single root occupies level 0, and each level below can have at most twice as many nodes as the level directly above it, since every node passes on at most 2 children to the next level. This gives a doubling (geometric) relationship between a level’s index and its maximum node count.

Application: Building up level by level from the root:

1. Level 0 (the root itself): at most 20 = 1 node.

2. Level 1: each of the root’s up to 2 children gives at most 2 × 1 = 21 = 2 nodes.

3. Level 2: each node at level 1 contributes at most 2 children, giving at most 2 × 2 = 22 = 4 nodes.

4. In general, level i has at most twice the maximum of level i-1, i.e. 2i nodes — this follows by induction on i.

5. Applying this to level L: the maximum number of nodes at level L is 2L.

Cross-check: Sum the per-level maximums for a perfect binary tree of height h (levels 0 through h): 20 + 21 + … + 2h = 2(h+1) - 1, which is exactly the well-known total-node-count formula for a perfect binary tree — confirming the per-level formula 2i is consistent.

Result: The maximum number of nodes at level L, with the root at level 0, is 2L.

Note: this question anchors the root explicitly at level 0 — the convention used by GATE-style/competitive DSA courses and most standard algorithms texts (e.g. CLRS). Some other textbooks number the root as level 1, under which the analogous maximum at level L would instead be 2(L-1); always confirm which convention a source uses before applying either formula.