LeetCode 124: Binary Tree Maximum Path Sum 2 I'm posting my code for a LeetCode problem. If you'd like to review, please do so. Thank you for your time!
Problem
Given a non-empty binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.
Example 1:
Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:
Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

Inputs
[1,2,3]
[-10,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7]
[-10,9,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7999999,20,null,null,15,7,9,20,null,null,15,720,null,null,15,7,9,20,null,null,15,7]

Outputs
6
42
66
791
8001552

Code
# 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) -&gt; int:
        def depth_first_search(node: TreeNode) -&gt; int:
            if not node:
                return 0
            left_sum = depth_first_search(node.left)
            right_sum = depth_first_search(node.right)
            if not left_sum or left_sum &lt; 0:
                left_sum = 0
            if not right_sum or right_sum &lt; 0:
                right_sum = 0
            self.sum = max(self.sum, left_sum + right_sum + node.val)
            return max(left_sum, right_sum) + node.val

        self.sum = float('-inf')
        depth_first_search(root)
        return self.sum


References

Problem

Discuss

Solution
Unnecessary checks in depth_first_search()
The function depth_first_search() always returns an integer value, never None. So the check for a partial sum being None or &lt; 0 can be rewritten using max():
left_sum = max(0, depth_first_search(node.left))
right_sum = max(0, depth_first_search(node.right))