题解 | #二叉树中的最大路径和#

题本身不难,但是被栈溢出卡了半天,python3的栈默认限制为1000,这也太小了吧.所以要先把recursion limit 设大一点. 之后就正常了.

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型
#
import sys
sys.setrecursionlimit(int(1e5)+100)
class Solution:
    def __init__(self):
        self.mx=int(-1e18)
        self.st=set()

    def dfs(self,cur: TreeNode):
        if cur==None:
            return 0
        lmx=self.dfs(cur.left)
        rmx=self.dfs(cur.right)
        self.mx=max(self.mx,lmx+rmx+cur.val)
        return max(0,cur.val+max(lmx,rmx))
        
    def maxPathSum(self , root: TreeNode) -> int:
        self.dfs(root)
        return self.mx