Leetcode\96. 不同的二叉搜索树\Python题解

class Solution:

    def numTrees(self, n: int) -> int:

        #dp[n]即n个节点可构成多少个二叉搜索树

        #动态转移dp[n] = (1)dp[0]*dp[n-1] + (2)dp[1]*dp[n-2] + ...(n)dp[n-1]*dp[0]

        dp = [1]*(n+1)

        for i in range(1,n+1):

            cur = 0

            for j in range(0,i):

                cur += dp[j]*dp[i-j-1]

            dp[i] = cur

        return dp[n]