https://leetcode.com/problems/unique-binary-search-trees-ii/

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

和上一篇类似

用分治的思想

每个点枚举作为根节点，递归枚举左子树，右子树

注意主函数判断0直接返回[] 否则返回的是[[]]

TreeNode* root=new TreeNode(i);为什么只能放在循环里啊……

/**
 * 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*> dfs(int l,int r){
        vector<TreeNode*>ans;
        if(l>r){
            ans.push_back(NULL);
            return ans;
        }
        for(int i=l;i<=r;i++){
            vector<TreeNode*>ansL=dfs(l,i-1);
            vector<TreeNode*>ansR=dfs(i+1,r);
            
            for(int j=0;j<ansL.size();j++){
                for(int k=0;k<ansR.size();k++){
                    TreeNode* root=new TreeNode(i);
                    root->left=ansL[j];
                    root->right=ansR[k];
                    ans.push_back(root);
                }
            }
        }
        return ans;
        
    }
    vector<TreeNode*> generateTrees(int n) {
        vector<TreeNode*> r;
        if(n==0)
            return r;
        return dfs(1,n);
    }
};