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The number of nodes per level grows as a power of 2 because each recursive call splits into exactly two subcalls. Starting with 1 node at the root (level 1, $2^0$), the second level has 2 nodes ($2^1$), the third has 4 nodes ($2^2$), and the fourth has 8 nodes ($2^3$).

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.

The number of nodes per level grows as a power of 2 because each recursive call splits into exactly two subcalls. Starting with 1 node at the root (level 1, $2^0$), the second level has 2 nodes ($2^1$), the third has 4 nodes ($2^2$), and the fourth has 8 nodes ($2^3$).

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.

The base case $f(1) = f(2) = 1$ determines where the recursion stops, meaning certain branches halt splitting at the bottom level. While this allows the instructor to omit drawing the entire tree in practice, the theoretical complexity analysis still estimates the scale based on the full binary tree's node count.

Conditions: The recurrence relation is $f(n) = f(n-1) + f(n-2)$.; The base cases are $f(1) = 1$ and $f(2) = 1$.; Complexity analysis uses a full binary tree for theoretical estimation.

The base case $f(1) = f(2) = 1$ determines where the recursion stops, meaning certain branches halt splitting at the bottom level. While this allows the instructor to omit drawing the entire tree in practice, the theoretical complexity analysis still estimates the scale based on the full binary tree's node count.

Conditions: The recurrence relation is $f(n) = f(n-1) + f(n-2)$.; The base cases are $f(1) = 1$ and $f(2) = 1$.; Complexity analysis uses a full binary tree for theoretical estimation.