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

由於1~n是升序列，因此建起來的樹天然就是BST。

遞迴思想，依次選擇根節點，對左右子序列再分別建樹。

由於左右子序列建樹的結果也可能不止一種，需要考慮所有搭配情況。

vector<TreeNode *> left代表所有valid左子樹。

vector<TreeNode *> right代表所有valid右子樹。

/**
 * 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 Helper(1, n);
    }
    vector<TreeNode *> Helper(int begin, int end)
    {
        vector<TreeNode *> ret;
        if(begin > end)
            ret.push_back(NULL);
        else if
 
(begin == end)
        {
            TreeNode* node = new TreeNode(begin);
            ret.push_back(node);
        }
        else
        {
            for(int i = begin; i <= end; i ++)
            {//root
                vector<TreeNode *> left = Helper(begin, i-1);
                vector<TreeNode *> right = Helper(i+1, end);
                for(int l = 0; l < left.size(); l ++)
                {
                    for(int r = 0; r < right.size(); r ++)
                    {
                        //new tree
                        TreeNode* root = new TreeNode(i);
                        root->left = left[l];
                        root->right = right[r];
                        ret.push_back(root);
                    }
                }
            }
        }
        return ret;
    }
};