## Unique Binary Search Trees II


### 描述

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
```


### 分析

见前面一题。


### 代码

{% if book.java %}
```java
// Unique Binary Search Trees II
// 时间复杂度TODO，空间复杂度TODO
public class Solution {
public List generateTrees(int n) {
if (n == 0) return new ArrayList<>();
return generate(1, n);
}
private static List generate(int start, int end) {
List subTree = new ArrayList<>();
if (start > end) {
subTree.add(null);
return subTree;
}
for (int k = start; k <= end; k++) {
List leftSubs = generate(start, k - 1);
List rightSubs = generate(k + 1, end);
for (TreeNode i : leftSubs) {
for (TreeNode j : rightSubs) {
TreeNode node = new TreeNode(k);
node.left = i;
node.right = j;
subTree.add(node);
}
}
}
return subTree;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Unique Binary Search Trees II
// 时间复杂度TODO，空间复杂度TODO
class Solution {
public:
vector generateTrees(int n) {
if (n == 0) return vector();
return generate(1, n);
}
private:
vector generate(int start, int end) {
vector subTree;
if (start > end) {
subTree.push_back(nullptr);
return subTree;
}
for (int k = start; k <= end; k++) {
vector leftSubs = generate(start, k - 1);
vector rightSubs = generate(k + 1, end);
for (auto i : leftSubs) {
for (auto j : rightSubs) {
TreeNode *node = new TreeNode(k);
node->left = i;
node->right = j;
subTree.push_back(node);
}
}
}
return subTree;
}
};
```
{% endif %}


### 相关题目


* [Unique Binary Search Trees](unique-binary-search-trees.md)