How does summing the number of nodes per level yield $O(2^n)$?

Conditions

• Time cost is measured by the number of recursive calls.
• An upper bound estimation using a perfect binary tree is adopted.
• The sum starts from $2^0$ with a common ratio of 2.

Reasoning, step by step

1. Count the number of nodes at each level as powers of 2 ($2^0, 2^1, 2^2, \dots$).
2. Sum the number of nodes across all levels to find the total size of the tree.
3. Use the geometric series example $2^0+2^1+\cdots+2^6=2^7-1$ to illustrate the growth trend.
4. Observe that the total sum is dominated by the highest power.
5. Conclude that the time complexity is $O(2^n)$.

Example

To illustrate the growth trend of such summations, he gives a concrete example: $2^0+2^1+\cdots+2^6=2^7-1$. This equation itself is a classic finite geometric series sum, and the video uses it to support the conclusion that when the number of levels grows with input size, the total number of nodes is dominated by the highest-order term, so the time complexity can be written as $O(2^n)$.

Common misconceptions

• Assuming the exact number of nodes is always $2^n - 1$ for any $n$, rather than using it as an asymptotic upper bound.
• Believing that the base case truncation reduces the theoretical exponential scale of the complexity.

Watch the explanation

Explore next

• Why does the video mention that 2 raised to the power 0 plus up to 2 raised to the power 6 equals 2 raised to the power 7 minus 1?
• What is $O(2^n)$ called in the video?
• What role does the base case $f(1) = f(2) = 1$ play in this analysis?