题目浅析

  • 想查看原题可以点击题目链接

  • 简单地说,就是找到二叉树中的一条数值最大的路径的数值和。

思路分享

代码解答(强烈建议自行解答后再看)

  • 参考题解
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# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxPathSum(self, root: Optional[TreeNode]) -> int:
ans = -1001
def dfs(root: Optional[TreeNode]) -> int:
if not root:
return 0
left = dfs(root.left)
right = dfs(root.right)
cur = root.val + left + right
nonlocal ans
if cur > ans:
ans = cur
return max(max(0, left, right) + root.val, 0)
dfs(root)
return ans

# 最初解法
ans = -1001
def dfs(root: Optional[TreeNode]) -> int:
if not root:
return 0
left = dfs(root.left)
right = dfs(root.right)
cur = root.val
if left > 0:
cur += left
if right > 0:
cur += right
nonlocal ans
if cur > ans:
ans = cur
return max(root.val, root.val + left, root.val + right)
dfs(root)
return ans