LeetCode "Unique Binary Search Trees"

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;
    }
};