96. Unique Binary Search Trees

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

class Solution:
    def numTrees(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = [0 for i in range(n+1)]
        res[0] = res[1] = 1
        for i in range(2,n+1):
            for j in range(i):
                res[i] += res[j]*res[i-j-1]
        return res[n]

卡罗兰公式

记G(n)为长度为n的数的个数，F(i,n)为以i为顶点,总顶点数为n的情况下解的个数.
则G(n) = F(1,n)+F(2,n)+...+F(n,n)
而 F(i,n)=G(i-1)*G(n-i) 1<=i<=n(由二叉搜索树的性质得出)
故
G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0)
参考自: https://blog.csdn.net/u012501459/article/details/46622501