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

题意和96题差不多，但是96题是问有多少种，而这一题是要求出这几种的BST。所以这一题会更难一点，但是还是可以通过递归很好地实现。基本思路都是一样的，先以某个点为根，然后分别递归地求出左子树和右子树，然后通过嵌套循环，得到一颗BST，代码如下。
Code:(LeetCode运行16ms)
/**
* 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) {
if (n == 0) {
vector<TreeNode *> result;
return result;
}
return generate(1, n);
}
private :
vector<TreeNode*> generate(int start, int end) {
vector<TreeNode *> result;
if (start > end) {
result.push_back(NULL);
return result;
}

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