# 124-binary-tree-maximum-path-sum


Try it on <a href='https://leetcode.com/problems/124-binary-tree-maximum-path-sum'>leetcode</a>

## Description
<div class="description">
<div><p>A <strong>path</strong> 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 <strong>at most once</strong>. Note that the path does not need to pass through the root.</p>

<p>The <strong>path sum</strong> of a path is the sum of the node's values in the path.</p>

<p>Given the <code>root</code> of a binary tree, return <em>the maximum <strong>path sum</strong> of any <strong>non-empty</strong> path</em>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg" style="width: 322px; height: 182px;">
<pre><strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The optimal path is 2 -&gt; 1 -&gt; 3 with a path sum of 2 + 1 + 3 = 6.
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg">
<pre><strong>Input:</strong> root = [-10,9,20,null,null,15,7]
<strong>Output:</strong> 42
<strong>Explanation:</strong> The optimal path is 15 -&gt; 20 -&gt; 7 with a path sum of 15 + 20 + 7 = 42.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 3 * 10<sup>4</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>
</div>
</div>

## Solution(Python)
```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:
        res = float('-inf')
        def dfs(node):
            nonlocal res;
            if not node:
                return 0
            
            left = max(dfs(node.left),0)
            right = max(dfs(node.right),0)
            
            cur_gain = node.val + max(left, right)
            res = max(res, node.val + left + right)
            return cur_gain
        dfs(root)
        return res
```