Unique Binary Search Trees

//Sigma(左边的子树可能状态 * 右边子树可能状态) = 当前个数的nodes形成的BST总数。
class Solution {
public:
    int numTrees(int n) {
        int *m = new int[n + 1]();
        m[0] = 1;
        m[1] = 1;
        for (int i = 2; i <= n; i++)
        {
            for (int j = 0; j < i; j++)
            {
                m[i] += m[j] * m[i -1 - j];
            }
        }
        int result = m[n];
        delete m;
        return result;
    }
};