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[Python手撕]不同的二叉搜索树

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

        dp = [0]*(n+1)
        dp[0] = 1
        dp[1] = 1

        for i in range(2,n+1):
            count = 0
            for left in range(0,i):
                count += (dp[left]*dp[i-left-1])
            dp[i] = count
        return dp[-1]
posted @ 2024-10-06 11:34  Duancf  阅读(10)  评论(0编辑  收藏  举报