<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:
    void dfs(int grand_parent, TreeNode* parent,vector<int>&vals){
        if (parent == nullptr) return;
        if (parent->left){
            if (grand_parent != -1 && grand_parent%2==0){
                vals.push_back(parent->left->val);
            }
            dfs(parent->val,parent->left,vals);
        }
        if (parent->right){
            if (grand_parent != -1 && grand_parent%2==0){
                vals.push_back(parent->right->val);
            }
            dfs(parent->val,parent->right,vals);
        }
        else{
            return;
        }
    }
    int sumEvenGrandparent(TreeNode* root) {

        if (root == NULL) return 0;
        vector<int>vals;
        dfs(-1,root,vals);
        int ans = accumulate(vals.begin(),vals.end(),0);
        return ans;
        
    }
};
40500