# 1315. Sum of Nodes with Even-Valued Grandparent

## Description

Given the root of a binary tree, return the sum of values of nodes with an even-valued grandparent. If there are no nodes with an even-valued grandparent, return 0.

A grandparent of a node is the parent of its parent if it exists.

Example 1:

Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output: 18
Explanation: The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.


Example 2:

Input: root = [1]
Output: 0


Constraints:

• The number of nodes in the tree is in the range [1, 104].
• 1 <= Node.val <= 100

## Solutions

DFS.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int res;

public int sumEvenGrandparent(TreeNode root) {
res = 0;
dfs(root, root.left);
dfs(root, root.right);
return res;
}

private void dfs(TreeNode g, TreeNode p) {
if (p == null) {
return;
}
if (g.val % 2 == 0) {
if (p.left != null) {
res += p.left.val;
}
if (p.right != null) {
res += p.right.val;
}
}
dfs(p, p.left);
dfs(p, p.right);
}
}

• /**
* 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 res;

int sumEvenGrandparent(TreeNode* root) {
res = 0;
dfs(root, root->left);
dfs(root, root->right);
return res;
}

void dfs(TreeNode* g, TreeNode* p) {
if (!p) return;
if (g->val % 2 == 0) {
if (p->left) res += p->left->val;
if (p->right) res += p->right->val;
}
dfs(p, p->left);
dfs(p, p->right);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def sumEvenGrandparent(self, root: TreeNode) -> int:
self.res = 0

def dfs(g, p):
if p is None:
return
if g.val % 2 == 0:
if p.left:
self.res += p.left.val
if p.right:
self.res += p.right.val
dfs(p, p.left)
dfs(p, p.right)

dfs(root, root.left)
dfs(root, root.right)
return self.res


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/

var res int

func sumEvenGrandparent(root *TreeNode) int {
res = 0
dfs(root, root.Left)
dfs(root, root.Right)
return res
}

func dfs(g, p *TreeNode) {
if p == nil {
return
}
if g.Val%2 == 0 {
if p.Left != nil {
res += p.Left.Val
}
if p.Right != nil {
res += p.Right.Val
}
}
dfs(p, p.Left)
dfs(p, p.Right)
}

• /**
* 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);
}