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:

   1
  / 
 2   3
    /
   4
    
     5

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

 

代码：

    vector<TreeNode *> generateTrees(int n, int start, int end){
        vector<TreeNode *> result;
        result.clear();
        if(n == 0){
            result.push_back(NULL);
            return result;
        }
        TreeNode *root;
        for(int i = start; i <= end; i++){
            vector<TreeNode *> l = generateTrees(i-start, start, i-1);
            vector<TreeNode *> r = generateTrees(end - i, i+1, end);
            int m = l.size(), n = r.size();
            int j,k;
            for(j = 0; j < m; j++){
                for(k = 0; k < n; k++){
                    root = new TreeNode(i);
                    root->left = l[j];
                    root->right = r[k];
                    result.push_back(root);
                }
            }
        }
        return result;
    }
    vector<TreeNode *> generateTrees(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        return generateTrees(n, 1, n);
    }