题目
给定一个非空二叉树，返回其最大路径和。
本题中，路径被定义为一条从树中任意节点出发，达到任意节点的序列。该路径至少包含一个节点，且不一定经过根节点。
示例 1:
输入: [1,2,3]
1
/
2 3
输出: 6
示例 2:
输入: [-10,9,20,null,null,15,7]
-10
/
9 20
/
15 7
输出: 42
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-maximum-path-sum
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路
- 树结构，遍历相关 ==> 递归思路实现。将边界条件设定好就行了。
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def maxSubSum(self, root):
if not root:
return -float('inf') , -float('inf')
lroot_subsum, lnonroot_subsum = self.maxSubSum(root.left)
rroot_subsum, rnonroot_subsum = self.maxSubSum(root.right)
return max(rroot_subsum +root.val, lroot_subsum +root.val, root.val), max(lroot_subsum + rroot_subsum +root.val, lnonroot_subsum, rnonroot_subsum, lroot_subsum, rroot_subsum)
def maxPathSum(self, root: TreeNode) -> int:
lroot_subsum, lnonroot_subsum = self.maxSubSum(root.left)
rroot_subsum, rnonroot_subsum = self.maxSubSum(root.right)
#print(lroot_subsum, lnonroot_subsum, rroot_subsum, rnonroot_subsum)
return max(lroot_subsum + rroot_subsum +root.val, rroot_subsum +root.val, lroot_subsum +root.val, root.val, lnonroot_subsum, rnonroot_subsum, lroot_subsum, rroot_subsum)