<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://assets.leetcode.com/uploads/2021/08/10/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://assets.leetcode.com/uploads/2021/08/10/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>

/**
 * 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 sumEvenGrandparent(TreeNode* root) {
        if(!root || (!root->left && !root->right)){
            return 0;
        }
        // Use of BFS
        int sum = 0;
        queue<pair<TreeNode*, pair<TreeNode*, TreeNode*>>> q;
        q.push({root, {NULL, NULL}});

        while(!q.empty()){            
            int size = q.size();

            for(int i = 0; i < size; i++){
                TreeNode *node = q.front().first;
                TreeNode *parent = q.front().second.first;
                TreeNode *gParent = q.front().second.second;
                q.pop();
                if(gParent && gParent->val % 2 == 0){
                    sum += node->val;
                }

                if(node->left){
                    q.push({node->left, {node, parent}});
                }
                if(node->right){
                    q.push({node->right, {node, parent}});
                }
            }
            
        }
        return sum;
    }
};
42400