124. Binary Tree Maximum Path Sum

dfs

Given anon-emptybinary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must containat least one nodeand does not need to go through the root.

Example 1:

Input:
 [1,2,3]


1
/ \
2
3
Output:
 6

Example 2:

Input:
 [-10,9,20,null,null,15,7]

   -10
   / \
  9  
20
/  \
15   7
Output:
 42

分析

无论任何情况， 必须要包含根节点！！！！ return = root+max(left,right), result = root+left+right

1. 处理负值：

   • 使用 max(dfs(node.left), 0) 确保不考虑负贡献

   • 这样能正确处理所有节点值为负的情况

2. 路径计算：

   • path_sum 计算当前节点作为最高点的完整路径和

   • res 记录全局最大值

3. 返回值：

   • 只能返回当前节点加上左或右子树的最大贡献值

   • 这样保证返回的是单边路径

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    ans = float("-inf")
    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """

        self.dfs(root)
        return self.ans
    def dfs(self,root):
        if not root:
            return 0
        left = max(0,self.dfs(root.left))
        right = max(0,self.dfs(root.right))
        self.ans = max(self.ans,left+right+root.val)
        return max(0,left+root.val,right+root.val)