Unique Binary Search Trees II

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.
For example,
Given n = 3, your program should return all 5 unique BST's shown below.
  1       3 　　 3 　　 2 　　 1
          / 　　 / 　　 / 　　
    3    2 　　 1 　　1  3 　　 2
   /     / 　　   　　　　　　　 
  2    1 　　　  2                    3

Solution: 1. DFS directly. (from the Internet)
2. DP + DFS. (my solution)
a. Generate trees for 'n' from 1 to n. (DP)
b. When generate trees for n = i, get the left and right subtrees
by copying tree structures of dp[1...i-1]. (copy tree uses DFS)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generateTrees(int n) {
        return generateTreesRe(1, n);
    }
    
    vector<TreeNode*> generateTreesRe(int l, int r)
    {
        vector<TreeNode*> res;
        if(l > r) {
            res.push_back(NULL);
            return res;
        }
        
        for(int k = l; k <= r; k++) {
            vector<TreeNode*> left = generateTreesRe(l, k-1);
            vector<TreeNode*> right = generateTreesRe(k+1, r);
            for(int i = 0; i < left.size(); i++) {
                for(int j = 0; j < right.size(); j++) {
                    TreeNode* root = new TreeNode(k);
                    root->left = left[i];
                    root->right = right[j];
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};