<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://assets.leetcode.com/uploads/2020/10/13/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://assets.leetcode.com/uploads/2020/10/13/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>

# 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:
    max_length: int = 0
    def longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
        def dfs(node: TreeNode):
            if not node:
                return (-1, -1)
            
            left, L_val = dfs(node.left)
            right, R_val = dfs(node.right)

            if node.val != L_val and node.val == R_val:
                right += 1
                self.max_length = max(self.max_length, right)
                return (right, node.val)
            elif node.val != R_val and node.val == L_val:
                left += 1
                self.max_length = max(self.max_length, left)
                return (left, node.val)
            elif node.val != L_val and node.val != R_val:
                return (0, node.val)
            else:
                left +=1
                right += 1
                self.max_length = max(self.max_length, left+right)
                return (max(left, right), node.val)
        dfs(root)
        return self.max_length
19400