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

题目大意很简单，就是计算有多少个不同的二叉寻找树满足前序遍历是1,2,3...n的条件

思路：每一棵树都是由根节点，左子树，右子树构成，一旦确定根节点是x的同时，

左子树只能由1,2,3...x-1构成，同理右子树由x+1,x+2...n构成，可以
[方法1]跌代求解；

class Solution:    # @return an integer    def numTrees(self, n):        dp = [1, 1, 2]        if n < 3: return dp[n]        dp += [0 for i in range(n-2)]        for i in range(3, n+1):            for j in range(i):                dp[i] += dp[j]*dp[i-j-1]        return dp[n]

[方法2]递归求解

class Solution:    # @return an integer    def numTrees(self, n):        dp = [1, 1, 2]        if n < 3: return dp[n]        ans = 0          for i in range(n):              ans += self.numTrees(i)*self.numTrees(n-i-1)          return ans

[leetcode] Unique Binary Search Trees @ Python