Unique Binary Search Trees ( I and II)

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

//Unique Binary Search Trees

class Solution {
    unordered_map<int, int> r;
public:
    int process(int n){
        if (r[n]) return r[n];
        int result = 0;
        for(int left = 0; left < n; ++left){
            result += process(left)*process(n-left-1);
        }
        r[n] = result;
        return result;
    }
    int numTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        r.clear();
        r[0] = 1;
        r[1] = 1;
        return process(n);
    }
};

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

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<TreeNode* > build(int start, int end){//inclusive
        if (start > end) return vector<TreeNode*>(1,(TreeNode*)NULL);
        if (start == end) return vector<TreeNode*>(1,new TreeNode(start));
        vector<TreeNode* > r;
        for(int root_val = start ;root_val <= end; ++root_val){
            vector<TreeNode* > r_left = build(start, root_val - 1);
            vector<TreeNode* > r_right = build(root_val+1, end);
            
            for(int ii = 0; ii < r_left.size(); ++ii){
                for(int jj = 0; jj < r_right.size(); ++jj){
                    TreeNode* head = new TreeNode(root_val);
                    head->left = r_left[ii];
                    head->right = r_right[jj];
                    r.push_back(head);
                }
            }
        }
        return r;
    }
public:
    vector<TreeNode *> generateTrees(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return build(1,n);
    }
};