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

思路：参照程序注释，写的很明白

#include <iostream>

class Solution {
public:
    int *dp;
    int myNumTreesMemorization(int n){
        if (dp[n]){
            return dp[n];
        }
        int & sum = dp[n];
        for(int i = 0; i < n; i++){
            dp[i] = myNumTreesMemorization(i);
            dp[n-1-i] = myNumTreesMemorization(n - 1 - i);
            sum += dp[i]*dp[n-1-i];
        }
        return sum;
    }
    
    
    /**
    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
    dp[0] = 1; dp[1] = 1;
    
    dp[n] =  1) n作为根结点，之前的二叉树就是left.child 这种情况： dp[n-1]*dp[0]
             2) n-1 是根结点，那么 n只有一种情况 ： dp[n-2] * dp[1]
             3) n-2 是根节点，right tree: dp[2] * left child : dp[n -3]
             ...
             4) 1 是根结点：dp[0] * dp[n-1]
    dp[n] = sigma{i = 0..n-1} dp[i]*dp[n-1-i]
    
    其实就是说白了考虑左右子tree的大小就ok，因为大小暗含了这个时候根结点的情况，所以和枚举根结点从1...n是一样的
    */
      
    int numTreesIteration(int n) {
        for(int i = 2; i <= n; i++){
            for(int j = 0; j < i; j++){
                dp[i] += dp[j]*dp[i-1-j];
            }
        }
        return dp[n];
    }

    int numTrees(int n){
        dp = new int[n + 1];
        memset(dp,0,sizeof(int)*(n + 1));
        dp[0] = dp[1] = 1;
        
        //int ans = numTreesIteration(n);
        int ans = myNumTreesMemorization(n);
        delete dp;
        return ans;
    }
};

using namespace std;
int main(int argc, char *argv[]) {
    int a = 4;
    Solution sol;
    cout << "solution: " << sol.numTrees(a) << endl;
}