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Formatted question description: https://leetcode.ca/all/1315.html

# 1315. Sum of Nodes with Even-Valued Grandparent (Medium)

Given a binary tree, return the sum of values of nodes with even-valued grandparent.  (A grandparent of a node is the parent of its parent, if it exists.)

If there are no nodes with an even-valued grandparent, return 0.

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.


Constraints:

• The number of nodes in the tree is between 1 and 10^4.
• The value of nodes is between 1 and 100.

Related Topics:
Tree, Depth-first Search

## Solution 1. Pre-order Traversal

// OJ: https://leetcode.com/problems/sum-of-nodes-with-even-valued-grandparent/
// Time: O(N)
// Space: O(N)
class Solution {
int ans = 0;
vector<TreeNode*> v;
void preorder(TreeNode *root, int level) {
if (!root) return;
if (v.size() <= level) v.push_back(root);
else v[level] = root;
if (level - 2 >= 0 && v[level - 2]->val % 2 == 0) ans += root->val;
preorder(root->left, level + 1);
preorder(root->right, level + 1);
}
public:
int sumEvenGrandparent(TreeNode* root) {
preorder(root, 0);
return ans;
}
};

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumEvenGrandparent(TreeNode root) {
int sum = 0;
nodeQueue.offer(root);
booleanQueue.offer(false);
while (!nodeQueue.isEmpty()) {
TreeNode node = nodeQueue.poll();
boolean isEven = booleanQueue.poll();
boolean curEven = node.val % 2 == 0;
TreeNode left = node.left, right = node.right;
if (left != null) {
if (isEven)
sum += left.val;
nodeQueue.offer(left);
booleanQueue.offer(curEven);
}
if (right != null) {
if (isEven)
sum += right.val;
nodeQueue.offer(right);
booleanQueue.offer(curEven);
}
}
return sum;
}
}

############

/**
* 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);
}
}

• // OJ: https://leetcode.com/problems/sum-of-nodes-with-even-valued-grandparent/
// Time: O(N)
// Space: O(N)
class Solution {
int ans = 0;
vector<TreeNode*> v;
void preorder(TreeNode *root, int level) {
if (!root) return;
if (v.size() <= level) v.push_back(root);
else v[level] = root;
if (level - 2 >= 0 && v[level - 2]->val % 2 == 0) ans += root->val;
preorder(root->left, level + 1);
preorder(root->right, level + 1);
}
public:
int sumEvenGrandparent(TreeNode* root) {
preorder(root, 0);
return ans;
}
};

• # 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)
}