LeetCode: 124. Binary Tree Maximum Path Sum

题目¶

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any path.

Example 1:

```Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
```

Example 2:

```Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
```

Constraints:

• The number of nodes in the tree is in the range [1, 3 * 10^4].
• -1000 <= Node.val <= 1000

解法¶

```# 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: TreeNode) -> int:
self._max_sum = -1001
self._path_sum(root)

return self._max_sum

def _path_sum(self, root):
if root is None:
return 0

left_sum = self._path_sum(root.left)
right_sum = self._path_sum(root.right)

# 如果子树的节点和 < 0，取 0 即舍弃这个子树
if left_sum < 0:
left_sum = 0
if right_sum < 0:
right_sum = 0

sum = root.val + left_sum + right_sum
if sum > self._max_sum:
self._max_sum = sum

return root.val + max(left_sum, right_sum)
```