Unique Binary Search Trees II -- leetcode

iven 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.

基本思路：递归

1. 依次以每个元素为根节点

2. 进行递归，该元素左边的所有元素，返回一组数的集合。

3. 该元素右边的所有元素，生成一组数的集合。

4. 分别从左边、右边集合各挑选一棵树作为左子树和右子树，连同根结点。组成一个棵树。

在leetcode上，实际执行时间为27ms。

/** * 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) {        vector<TreeNode *> ans;        generateTrees(ans, 1, n+1);        return ans;    }        void generateTrees(vector<TreeNode *> &ans, int start, int stop) {        for (int i=start; i<stop; i++) {            vector<TreeNode *> lefts;            generateTrees(lefts, start, i);            for (int j=0; j<lefts.size(); j++) {                vector<TreeNode *> rights;                generateTrees(rights, i+1, stop);                for (int k=0; k<rights.size(); k++) {                    TreeNode *root = new TreeNode(i);                    root->left = lefts[j];                    root->right = rights[k];                    ans.push_back(root);                }            }        }                if (ans.empty())            ans.push_back(0);    }};