给定结点数n，结点值为1，2，...，n，求由这些结点可以构成的所有二叉查找树。

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

思路：递归构造，分别构造出左，右子树，然后组合成来。

/**

 * 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 *> generateTrees(int n)

	{

        return generateTrees(1,n);

    }

	vector<TreeNode *> generateTrees(int start, int end)

	{

		vector<TreeNode *> trees;

		if (start > end)

		{

			trees.push_back(NULL);

			return trees;

		}

		if (start==end)

		{

			trees.push_back(new TreeNode(start));

			return trees;

		}

		for (int i=start; i<=end; ++i)

		{

			vector<TreeNode *> treesleft = generateTrees(start,i-1);

			vector<TreeNode *> treesright = generateTrees(i+1,end);

			for (size_t j=0; j<treesleft.size(); ++j)

			{

				for (size_t k=0; k<treesright.size(); ++k)

				{

					TreeNode *root = new TreeNode(i);

					root->left = treesleft[j];

					root->right = treesright[k];

					trees.push_back(root);

				}

			}

		}

		return trees;

	}

};