leetcode Unique Binary Search Trees

Given n, how many structurally unique BST’s (binary search trees) that store values 1…n?

For example,
Given n = 3, there are a total of 5 unique BST’s.

1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
题意：给定一个整数，从1~n可以组成多少个二叉搜索树。
思路：动态规划，当n=0或n=1时，共可构成1个二叉搜索树；当n>1时，取i（1<=i<=n）作为根节点，i左边数作为左子树，i右边数作为右子树，即共可构成dp[n] = dp[i-1]*dp[n-i]个二叉树。

class Solution {public:    int numTrees(int n) {        int *dp = new int[n+1];        dp[0] = dp[1] = 1;        for (int i = 2; i <= n; ++i) {            dp[i] = 0;            for (int j = 1; j <= i; ++j)                dp[i] += dp[j-1] * dp[i-j];        }        return dp[n];    }};