# [Solved] 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. ## Question

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`

## Python 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
max_sum = -2**31
def DFS(root):
global max_sum
if not root: return -2**31
l = DFS(root.left)
r = DFS(root.right)
v = root.val
max_sum = max(v,v+l,v+r,v+l+r,max_sum)
return max(l+v,r+v,v)

class Solution:
def maxPathSum(self, root: TreeNode) -> int:
global max_sum
max_sum = -2**31
DFS(root)
return max_sum```