Unique Binary Search Trees II

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

代码：

 1     vector<TreeNode *> generateTrees(int n, int start, int end){
 2         vector<TreeNode *> result;
 3         result.clear();
 4         if(n == 0){
 5             result.push_back(NULL);
 6             return result;
 7         }
 8         TreeNode *root;
 9         for(int i = start; i <= end; i++){
10             vector<TreeNode *> l = generateTrees(i-start, start, i-1);
11             vector<TreeNode *> r = generateTrees(end - i, i+1, end);
12             int m = l.size(), n = r.size();
13             int j,k;
14             for(j = 0; j < m; j++){
15                 for(k = 0; k < n; k++){
16                     root = new TreeNode(i);
17                     root->left = l[j];
18                     root->right = r[k];
19                     result.push_back(root);
20                 }
21             }
22         }
23         return result;
24     }
25     vector<TreeNode *> generateTrees(int n) {
26         // IMPORTANT: Please reset any member data you declared, as
27         // the same Solution instance will be reused for each test case.
28         return generateTrees(n, 1, n);
29     }