## Unique Binary Search Trees

### 描述

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.


1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3


### 分析


1 1 2 3 3
\ \ / \ / /
3 2 1 3 2 1
/ \ / \
2 3 1 2


\textbf{以i为根节点的树，其左子树由[1, i-1]构成， 其右子树由[i+1, n]构成。}


1 2
\ /
2 1


$$f(2) = f(0) * f(1) \text{ , when 1 as root}$$

$$+ f(1) * f(0) \text{ , when 2 as root}$$

$$f(3) = f(0) * f(2) \text{ , when 1 as root}$$

$$+ f(1) * f(1) \text{ , when 2 as root}$$

$$+ f(2) * f(0) \text{ , when 3 as root}$$

$$f(i) = \sum_{k=1}^{i} f(k-1) \times f(i-k)$$

### 代码

{% if book.java %}
{% codesnippet "./code/unique-binary-search-trees."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
cpp
// Unique Binary Search Trees
// 时间复杂度O(n^2)，空间复杂度O(n)
class Solution {
public:
int numTrees(int n) {
vector f(n + 1, 0);

f[0] = 1;
f[1] = 1;
for (int i = 2; i <= n; ++i) {
for (int k = 1; k <= i; ++k)
f[i] += f[k-1] * f[i - k];
}

return f[n];
}
};

{% endif %}

### 相关题目

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