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.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var longestUnivaluePath = function (root) {
    let ans = 0;
    const dfs = root => {
        if (!root) {
            return 0;
        }
        let [l, r] = [dfs(root.left), 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 = Math.max(ans, l + r);
        return Math.max(l, r);
    };
    dfs(root);
    return ans;
};
