95. Unique Binary Search Trees II

Given an integer 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

解题思路：
类似96. Unique Binary Search Trees的思路，二叉搜索树（BST）的建树原则如下：
以i为跟根节点的树，其左子树由[1,i-1]构成，右子树由[i+1, n]构成。
由于要输出BST序列，借鉴动态规划的做法，对从1～n的每个序列都进行建树：
1. 当子树的节点值比根节点值小的时候，该子树为空。
2. 递归建立左右子树，根节点连接左右子树。

/** * 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*> generateTrees(int n) {        vector<TreeNode *> result;        if (n == 0) return result;        return generateTrees(1, n);    }    vector<TreeNode*> generateTrees(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 *> leftSubs = generateTrees(start, k - 1);            vector<TreeNode *> rightSubs = generateTrees(k + 1, end);            for (auto i : leftSubs) {                for (auto j : rightSubs) {                    TreeNode *node = new TreeNode(k);                    node->left = i;                    node->right = j;                    result.push_back(node);                }            }        }        return result;    }};