95. Unique Binary Search Trees II

/*Given an integer 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                 3二叉查找树：    左节点小于父节点，父节点小于右节点*//** * 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) {        vector<TreeNode*> res;        if(n<=0) return res;        return generate(1,n);    }    vector<TreeNode*> generate(int begin,int end)    {        vector<TreeNode*> res;        if(begin>end)        {            res.push_back(NULL);            return res;        }        for(int i=begin;i<=end;i++)        {            vector<TreeNode*> leftTree=generate(begin,i-1);            vector<TreeNode*> rightTree=generate(i+1,end);            for(int j=0;j<leftTree.size();j++)                for(int k=0;k<rightTree.size();k++)                {                    TreeNode* father=new TreeNode(i);                    res.push_back(father);                    father->left=leftTree[j];                    father->right=rightTree[k];                }        }        return res;    }};