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 non-empty 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 * 104]
. -
-1000 <= Node.val <= 1000
SOLUTION:
# 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 pathsum(self, root):
if root:
l = self.pathsum(root.left)
r = self.pathsum(root.right)
self.msum = max(self.msum, root.val, l + root.val, root.val + r, l + root.val + r)
return max(root.val, root.val + l, root.val + r)
return float('-inf')
def maxPathSum(self, root: Optional[TreeNode]) -> int:
self.msum = float('-inf')
self.pathsum(root)
return self.msum
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