Count Univalue Subtrees
<p>Given the <code>root</code> of a binary tree, return <em>the number of <strong>uni-value</strong> </em><span data-keyword="subtree"><em>subtrees</em></span>.</p>

<p>A <strong>uni-value subtree</strong> means all nodes of the subtree have the same value.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0250.Count%20Univalue%20Subtrees/images/unival_e1.jpg" style="width: 450px; height: 258px;" />
<pre>
<strong>Input:</strong> root = [5,1,5,5,5,null,5]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [5,5,5,5,5,null,5]
<strong>Output:</strong> 6
</pre>

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

<ul>
	<li>The number of the node in the tree will be in the range <code>[0, 1000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>
Tree; Depth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function countUnivalSubtrees(root: TreeNode | null): number {
    let ans: number = 0;
    const dfs = (root: TreeNode | null): boolean => {
        if (root == null) {
            return true;
        }
        const l: boolean = dfs(root.left);
        const r: boolean = dfs(root.right);
        if (!l || !r) {
            return false;
        }
        if (root.left != null && root.left.val != root.val) {
            return false;
        }
        if (root.right != null && root.right.val != root.val) {
            return false;
        }
        ++ans;
        return true;
    };
    dfs(root);
    return ans;
}
