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Naive recursion is highly inefficient because it generates exponentially many repeated subproblems. When visualized as a recursion tree, identical nodes such as $f(4)$, $f(3)$, and $f(2)$ are triggered multiple times by different parent nodes, meaning the same calculations are redundantly executed across different branches.

Conditions: The algorithm uses naive recursion without memoization.; The recurrence relation is $f(n) = f(n-1) + f(n-2)$.

Naive recursion is highly inefficient because it generates exponentially many repeated subproblems. When visualized as a recursion tree, identical nodes such as $f(4)$, $f(3)$, and $f(2)$ are triggered multiple times by different parent nodes, meaning the same calculations are redundantly executed across different branches.

Conditions: The algorithm uses naive recursion without memoization.; The recurrence relation is $f(n) = f(n-1) + f(n-2)$.

To analyze the time complexity, the video treats each node in the recursion tree as one unit of computation (one function call). The total execution cost of the algorithm is then evaluated by counting the total number of nodes in the tree, including both internal nodes and leaf nodes.

Conditions: The recursive algorithm can be clearly represented by a recursion tree.; The overhead of each function call (excluding recursive calls) is assumed to be constant.

To analyze the time complexity, the video treats each node in the recursion tree as one unit of computation (one function call). The total execution cost of the algorithm is then evaluated by counting the total number of nodes in the tree, including both internal nodes and leaf nodes.

Conditions: The recursive algorithm can be clearly represented by a recursion tree.; The overhead of each function call (excluding recursive calls) is assumed to be constant.

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.

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.