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.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int longestUnivaluePath(TreeNode* root) {
        int ans = 0;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left);
            int r = dfs(root->right);
            l = root->left && root->left->val == root->val ? l + 1 : 0;
            r = root->right && root->right->val == root->val ? r + 1 : 0;
            ans = max(ans, l + r);
            return max(l, r);
        };
        dfs(root);
        return ans;
    }
};
