Binary Tree Maximum Path Sum
<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&#39;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 class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/images/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 class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/images/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>
Tree; Depth-First Search; Dynamic Programming; Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxPathSum(root *TreeNode) int {
	ans := -1001
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		left := max(0, dfs(root.Left))
		right := max(0, dfs(root.Right))
		ans = max(ans, left+right+root.Val)
		return max(left, right) + root.Val
	}
	dfs(root)
	return ans
}