Why is the number of nodes per level a power of 2?

Conditions

• The recurrence relation is $f(n) = f(n-1) + f(n-2)$, producing a binary tree.
• Levels are numbered starting from 1.
• The tree is counted as a full binary tree for theoretical estimation.

Reasoning, step by step

1. Identify the root node as level 1, containing 1 node ($2^0$).
2. Observe that each node splits into two child nodes in the next level.
3. Count the nodes at level 2 as 2 ($2^1$), level 3 as 4 ($2^2$), and level 4 as 8 ($2^3$).
4. Generalize the pattern to conclude that level $k$ corresponds to $2^{k-1}$ nodes.

Example

The instructor points layer by layer from top to bottom: the second level has 2 nodes, the third level has 4 nodes, and the fourth level has 8 nodes, writing $2^0$, $2^1$, $2^2$, and $2^3$ beside the tree in sequence. The mathematical reasoning is straightforward: each recursive call splits into two subcalls, so the number of nodes per level grows as a power of 2.

Common misconceptions

• Assuming the number of nodes per level is constant across the entire tree.
• Confusing the level index $k$ with the exponent, forgetting that level 1 corresponds to $2^0$.

Watch the explanation

Explore next

• How does summing the number of nodes per level yield $O(2^n)$?
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• What is the time complexity of this recursive algorithm?