【Leetcod】Unique Binary Search Trees II

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;
	}
};