96. Unique Binary Search Trees

题目链接
Unique Binary Search Trees

题目描述
Given n, how many structurally unique BST’s (binary search trees) that store values 1 … n?

Example:

Input: 3
Output: 5
Explanation:
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

解题思路：

刚拿到这个题目时，不知道怎么做，于是想到了递归，假设现在n=5，那么这5个元素每个都有可能是根节点，可以发现规律如下：

于是就有
f(n) = f(0)*f(n-1) + f(1)*f(n-2) + f(2)*f(n-3) + … + f(n-1) *f(0)

因此代码如下所示：

class Solution {
    public int numTrees(int n) {
        if(n == 0 || n == 1) {
            return 1;
        }
        int[] nums = new int[n+1];
        nums[0] = 1;
        nums[1] = 1;
        nums[2] = 2;
        if(n <= 2) {
            return nums[n];
        }
        for(int i = 3; i <=n; i++) {
            int sum = 0;
            for(int j = 0; j < i; j++) {
                sum += nums[j] * nums[i - 1 - j];
            } 
            nums[i] = sum;
        }
        return nums[n];
    }
}