LeetCode95 Unique Binary Search Trees II

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

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

分析：

这个题比前一道难在要构造出具体的树。但基本思路还是一样，枚举根节点，递归调用生成左右子树，然后左右子树结果组合在一个得到所有结果。

代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 private:
12     vector<TreeNode*> helper(int start, int end) {
13         if (start > end) {
14             return vector<TreeNode*>{nullptr};
15         }
16         if (start == end) {
17             TreeNode* root = new TreeNode(start);
18             return vector<TreeNode*>{root};
19         }
20         vector<TreeNode*> result;
21         for (int i = start; i <= end; ++i) {
22             vector<TreeNode*> leftV = helper(start, i - 1);
23             vector<TreeNode*> rightV = helper(i + 1, end);
24             for (int j = 0; j < leftV.size(); ++j) {
25                 for (int k = 0; k < rightV.size(); ++k) {
26                     TreeNode* root = new TreeNode(i);
27                     root -> left = leftV[j];
28                     root -> right = rightV[k];
29                     result.push_back(root);
30                 }
31             }
32         }
33         return result;
34     }
35 public:
36     vector<TreeNode*> generateTrees(int n) {
37         vector<TreeNode*> result;
38         if (n == 0) {
39             return result;
40         }
41         result = helper(1, n);
42         return result;
43     }
44 };