Another recursion-style problem.
1. Each count of sub-solution with a certain root should contribute to final count
2. With a certain root, the num of left child sub-tree multiply the right one, is the current count
1A, YEAH!
class Solution { public: int numTrees(int n) { if (n <= 1) return 1; if (n == 2) return 2; int cnt = 0; for (int i = 1; i <= n; i++) { int r1 = numTrees(i - 1); int r2 = numTrees(n - i); cnt += r1 * r2; } return cnt; } };
LeetCode "Unique Binary Search Trees",布布扣,bubuko.com
LeetCode "Unique Binary Search Trees"
原文：http://www.cnblogs.com/tonix/p/3853414.html