LSGO——LeetCode實戰（樹系列）：124題 二叉樹中的最大路徑和(Binary Tree Maximum Path Sum)

給定一個非空二叉樹，返回其最大路徑和。

本題中，路徑被定義爲一條從樹中任意節點出發，達到任意節點的序列。該路徑至少包含一個節點，且不一定經過根節點。

示例 1:

輸入: [1,2,3]

   1
  / \
 2   3

輸出: 6
示例 2:

輸入: [-10,9,20,null,null,15,7]

-10
/
9 20
/
15 7

輸出: 42

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/binary-tree-maximum-path-sum
著作權歸領釦網絡所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。

解

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

class Solution:
    def maxPathSum(self, root: TreeNode) -> int:
        def max_gain(node):
            nonlocal max_sum
            if not node:
                return 0

            # max sum on the left and right sub-trees of node
            left_gain = max(max_gain(node.left), 0)
            right_gain = max(max_gain(node.right), 0)
            
            # the price to start a new path where `node` is a highest node
            price_newpath = node.val + left_gain + right_gain
            
            # update max_sum if it's better to start a new path
            max_sum = max(max_sum, price_newpath)
        
            # for recursion :
            # return the max gain if continue the same path
            return node.val + max(left_gain, right_gain)
   
        max_sum = float('-inf')
        max_gain(root)
        return max_sum