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}".

思路：列出所有的二叉搜索树，只能是枚举了，而且是递归枚举；故使用DFS来枚举出所有的可能。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode *> generateBST(int left,int right)
    {
        vector<TreeNode *> result;
        if(left>right)
        {
            result.push_back(NULL);
            return result;
        }
        for(int i=left;i<=right;i++)
        {
            vector<TreeNode *> leftTree=generateBST(left,i-1);
            vector<TreeNode *> rightTree=generateBST(i+1,right);
            for(int j=0;j<leftTree.size();j++)
            {
                for(int k=0;k<rightTree.size();k++)
                {
                    TreeNode *root=new TreeNode(i);
                    result.push_back(root);
                    root->left=leftTree[j];
                    root->right=rightTree[k];
                }
            }
        }
        return result;
    }
    vector<TreeNode *> generateTrees(int n) {
        vector<TreeNode *> result;
        result=generateBST(1,n);
        return result;
    }
};