95.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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：DFS。取i作为根节点，递归求1->i-1和 i+1->n的二叉搜索树，置于 vector<TreeNode *> left 和vector<TreeNode *> right 之中，然后取i作为根节点，组合数组left和right，分别取一个元素作为root的左右子树，然后将组合之后的树置于result之中，最后返回result即可。注意一个特殊情况，当start>end,将空节点写入result，返回result，代表返回一个空节点。还有就是初始n<1的特殊处理。

/**
 * Definition for a binary tree node.
 * 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) {
        if(n<1){
            vector<TreeNode *> result;
            return result;
        }
        return help(1,n);
    }
    vector<TreeNode *> help(int start,int end){
        vector<TreeNode *> result;
        if(start>end){
            result.push_back(NULL);
            return result;
        }
        for(int i=start;i<=end;i++){
            vector<TreeNode *> left =help(start,i-1);
            vector<TreeNode *> right=help(i+1,end);
            for(int j=0;j<left.size();j++){
                for(int k=0;k<right.size();k++){
                    TreeNode *root =new TreeNode(i);
                    root->left=left[j];
                    root->right=right[k];
                    result.push_back(root);
                    
                }
            }
        }
        return result;
        
        
    }
};