关于递归式T(n)=T(n/2)+T(n/3)+1的递归树高度与末层节点数的技术咨询

Great question—since you already grasp the recursion tree for the version with n as the additive cost, let's unpack this simpler (but tricky!) case where we add 1 instead.

Recursion Tree Height #

First, let's clarify what we mean by "height": we're talking about the longest path from the root node (T(n)) to any leaf node (where recursion stops, typically when the input drops below 1, since we can't recurse on a fraction of a problem in practice).

Each node splits into two children: one for T(n/2) and one for T(n/3). The longest path will always follow the T(n/2) branch—because dividing by 2 shrinks the problem size more slowly than dividing by 3, so it takes more steps to reach the base case.

To calculate the exact height: we need the smallest number of steps k such that n/(2^k) ≤ 1. Solving for k, that's k = ⌈log₂n⌉. Let's use a concrete example to avoid confusion:

• Take n=6. The longest path from root to leaf is: T(6) → T(3) → T(1.5) → T(0.75) (we stop here because 0.75 < 1).
  • If we count edges (the number of splits), that's 3 edges—so height is 3.
  • If we count layers (root is layer 0), the deepest leaf is in layer 3—so height (max layer index) is 3.

Asymptotically, we can just say the height is Θ(log n)—since log base 2 and log base 3 are within a constant factor of each other, the exact base doesn't matter for big-O notation.

Number of Nodes in the Last Layer #

The last layer consists of all leaf nodes where recursion stops (input ≤1). Each leaf corresponds to a unique path from the root, where each step we chose either the T(n/2) or T(n/3) branch.

To find how many such leaves exist, think of each leaf as a sequence of choices that leads to n/(2^a * 3^b) ≤1 (where a is the number of times we took the T(n/2) branch, b the number of times we took T(n/3)). Rearranged, that means 2^a *3^b ≥n.

Asymptotically, this ties directly to the solution of the recurrence relation. The recurrence T(n) = T(n/2)+T(n/3)+1 has a solution of Θ(n^c), where c is the unique positive real number satisfying 1 = (1/2)^c + (1/3)^c (this is the characteristic equation for divide-and-conquer recurrences). Calculating c gives approximately 0.787. Since each leaf contributes exactly 1 to the total cost (the +1 in the recurrence), the total number of leaves is asymptotically Θ(n^c).

For a small example, take n=6:

• The leaves are T(0.75), T(1), T(1), T(0.666...)—that's 4 leaves. For n=6, n^c ≈6^0.787≈3.8, which is very close to the actual count of 4, so the asymptotic estimate holds even for small values of n.

内容的提问来源于stack exchange，提问作者spoph