<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:
    def __init__(self):
        self.ans  = 0
    
    def inorder(self, root, parent):
        if root is None:
            return 0
        
        lt = self.inorder(root.left, root.val)
        rt = self.inorder(root.right, root.val)
        
        self.ans = max(self.ans, lt+rt)
        
        return max(lt, rt) + 1 if root.val == parent else 0
        
        
    def longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
        self.inorder(root, -1)
        return self.ans
19300