# [leetcode] Unique Binary Search Trees II

### 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.

```//递归
//对于n个node，每次取1~n之间的一个数i作为root节点，那么它的left child便是1~i-1的所有可能组合，right child就是i+1~n的所有可能组合。
//_generateTrees(int start, int end)便是求出[start, end)(不包括end，包括start)的所有可能的树的情形。
//时间复杂度不清楚，但肯定是有冗余的。因为很多计算是不需要的。
/**
* 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) {
return _generateTree(1, n + 1);
}
//[start, end)
vector<TreeNode *> _generateTree(int start, int end){
vector<TreeNode *> re;
if(start >= end){
re.push_back(NULL);
return re;
}
else if(start + 1 == end){
re.push_back(new TreeNode(start));
return re;
}
else{
for(int i = start; i < end; i++){
vector<TreeNode*> left = _generateTree(start, i);
vector<TreeNode*> right = _generateTree(i + 1, end);
for(vector<TreeNode*>::iterator it1 = left.begin(); it1 != left.end(); it1++){
for(vector<TreeNode*>::iterator it2 = right.begin(); it2 != right.end(); it2++){
TreeNode * node = new TreeNode(i);
node->left = *it1;
node->right = *it2;
re.push_back(node);
}
}
}
return re;
}
}
};```

This site uses Akismet to reduce spam. Learn how your comment data is processed.