Sum of Nodes with Even-Valued Grandparent
<p>Given the <code>root</code> of a binary tree, return <em>the sum of values of nodes with an <strong>even-valued grandparent</strong></em>. If there are no nodes with an <strong>even-valued grandparent</strong>, return <code>0</code>.</p>

<p>A <strong>grandparent</strong> of a node is the parent of its parent if it exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1315.Sum%20of%20Nodes%20with%20Even-Valued%20Grandparent/images/even1-tree.jpg" style="width: 504px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
<strong>Output:</strong> 18
<strong>Explanation:</strong> The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1315.Sum%20of%20Nodes%20with%20Even-Valued%20Grandparent/images/even2-tree.jpg" style="width: 64px; height: 65px;" />
<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
</ul>
Tree; Depth-First Search; Breadth-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 sumEvenGrandparent(root: TreeNode | null): number {
    const dfs = (root: TreeNode | null, x: number): number => {
        if (!root) {
            return 0;
        }
        const { val, left, right } = root;
        let ans = dfs(left, val) + dfs(right, val);
        if (x % 2 === 0) {
            ans += left?.val ?? 0;
            ans += right?.val ?? 0;
        }
        return ans;
    };
    return dfs(root, 1);
}
