LeetCode 96:Unique Binary Search Trees
2016-01-04 09:23:29      个评论    来源：geekmanong的专栏

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
```

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```//由1,2,3,...,n构建的二叉查找树，以i为根节点，左子树由[1,i-1]构成，其右子树由[i+1,n]构成。
//定义f(i)为以[1,i]能产生的Unique Binary Search Tree的数目
//若数组为空，则只有一种BST，即空树，f(0)=1;
//若数组仅有一个元素1，则只有一种BST，单个节点，f(1)=1;
//若数组有两个元素1，2，则有两种可能，f(2)=f(0)*f(1)+f(1)*f(0);
//若数组有三个元素1，2，3，则有f(3)=f(0)*f(2)+f(1)*f(1)+f(2)*f(0)
//由此可以得到递推公式：f(i)=f(0)*f(i-1)+...+f(k-1)*f(i-k)+...+f(i-1)*f(0)
//利用一维动态规划来求解
class Solution {
public:
int numTrees(int n) {
vector f(n+1,0); //n+1个int型元素，每个都初始化为0
f[0] = 1;
f[1] = 1;
for (int i = 2; i <= n; ++i){
for (int k = 1; k <= i; ++k)
f[i] = f[i] + f[k - 1] * f[i - k];
}
return f[n];
}
};```