Unique Binary Search Trees II

题目：

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.

题目：

这题要求将每个子树输出，思想就是对于n，以1-n为根节点，然后左右子树分别是比其小和比其大的节点构成的，对左右子树的的形态也要使用遍历，代码中使用的是foreach遍历：for(i:j)这种形式

 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 public:
12     vector<TreeNode *> generateTrees(int n) {
13         if (n == 0) return generate(1, 0);
14         return generate(1, n);
15     }
16 private:
17     vector<TreeNode *> generate(int start, int end) {
18         vector<TreeNode*> subTree;
19         if (start > end) {
20             subTree.push_back(nullptr);
21             return subTree;
22         }
23         for (int k = start; k <= end; k++) {
24             vector<TreeNode*> leftSubs = generate(start, k - 1);
25             vector<TreeNode*> rightSubs = generate(k + 1, end);
26             for (auto i : leftSubs) {
27                 for (auto j : rightSubs) {
28                     TreeNode *node = new TreeNode(k);
29                     node->left = i;
30                     node->right = j;
31                     subTree.push_back(node);
32                 }
33             }
34         }
35         return subTree;
36     }
37 };