[Notes] 2020.6.21 每日一题 二叉树中的最大路径和

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

1. 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)