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.

题解：
这道题比1难的就是不是返回个数，而是返回全部结果。从http://blog.csdn.net/sinat_24520925/article/details/45562273可知可行的二叉查找树的数量时对应的卡特兰数。由于左右子数也是一棵树，仅仅是树节点的值不同，所以我们用递归的方法求全部树的结构。分成左子树以及右子树，将1~n先分成两部分，之后将这两部分递归所组成子树挂在根节点下就可以，代码例如以下：

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