#+PROPERTY: STUDY_DECK_02
* TODO 0124. Binary Tree Maximum Path Sum :hard:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0124. Binary Tree Maximum Path Sum][0124. Binary Tree Maximum Path Sum]]
:END:

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:*


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


*Example 2:*


#+begin_src
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.
#+end_src


*Constraints:*

 - The number of nodes in the tree is in the range ~[1, 3 * 10^{4}]~.

 - ~-1000 <= Node.val <= 1000~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python
# 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: Optional[TreeNode]) -> int:
#+end_src

** TODO C++
#+begin_src cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        
    }
};
#+end_src