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，则仅仅有一种BST，即空树。f(0)=1;//若数组仅有一个元素1，则仅仅有一种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)//由此能够得到递推公式：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。每一个都初始化为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];	}};