7/15刷题

1.不同的二叉搜索树

思路一：动态规划

class Solution {
public:
    int numTrees(int n) {
        vector<int> G(n + 1, 0);
        G[0] = 1;//无节点
        G[1] = 1;//一个节点

        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= i; ++j) {
                G[i] += G[j - 1] * G[i - j];//状态转移方程
            }
        }
        return G[n];
    }
};

思路二：数学

class Solution {
public:
    int numTrees(int n) {
        long long C = 1;
        for (int i = 0; i < n; ++i) {
            C = C * 2 * (2 * i + 1) / (i + 2);
        }
        return (int)C;
    }
};