Longest Univalue Path
<p>Given the <code>root</code> of a binary tree, return <em>the length of the longest path, where each node in the path has the same value</em>. This path may or may not pass through the root.</p>

<p><strong>The length of the path</strong> between two nodes is represented by the number of edges between them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0687.Longest%20Univalue%20Path/images/ex1.jpg" style="width: 450px; height: 238px;" />
<pre>
<strong>Input:</strong> root = [5,4,5,1,1,null,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The shown image shows that the longest path of the same value (i.e. 5).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0687.Longest%20Univalue%20Path/images/ex2.jpg" style="width: 450px; height: 238px;" />
<pre>
<strong>Input:</strong> root = [1,4,5,4,4,null,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The shown image shows that the longest path of the same value (i.e. 4).
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
	<li>The depth of the tree will not exceed <code>1000</code>.</li>
</ul>
Tree; Depth-First Search; Binary Tree
# 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 longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            l = l + 1 if root.left and root.left.val == root.val else 0
            r = r + 1 if root.right and root.right.val == root.val else 0
            nonlocal ans
            ans = max(ans, l + r)
            return max(l, r)

        ans = 0
        dfs(root)
        return ans
