Unique Binary Search Trees II(leetcode)

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.

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}"

.
解题思路：递归，参考了leetcode题解。调试时不需要把信息打印输出，只需要在return result处加一个断点，就能查看result的值了。这样能够避免写过多的调试代码。
代码：

#include<iostream>
#include<vector>
using namespace std;

struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

vector<TreeNode *> generateTrees(int n)
{
if(n==0)
return generate(1,0);
return generate(1,n);
}

vector<TreeNode *> generate(int start,int end)
{
vector<TreeNode *> result;
if(start>end)
{
result.push_back(nullptr);
return result;
}
for(int k=start;k<=end;k++)
{
vector<TreeNode*> succLeft=generate(start,k-1);
vector<TreeNode*> succRight=generate(k+1,end);
for(auto i=succLeft.begin();i!=succLeft.end();i++)
{
for(auto j=succRight.begin();j!=succRight.end();j++)
{
TreeNode *node=new TreeNode(k);
node->left=*i;
node->right=*j;
result.push_back(node);
}
}
}
return result;
}

int main()
{
generateTrees(5);

system("pause");
return 0;
}