Sum of Nodes with Even-Valued Grandparent

Sum of Nodes with Even-Valued Grandparent - Problem

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. Think of it as traversing three generations in the tree: grandparent → parent → grandchild.

Example: If a node has value 6 (even), then all of its grandchildren's values should be included in our sum. We need to look two levels down from each even-valued node.

This problem tests your ability to track relationships across multiple levels of a tree structure while traversing efficiently.

Input & Output

example_1.py — Standard Binary Tree

$ Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]

› Output: 18

💡 Note: Node 6 is the only even-valued grandparent. Its grandchildren are nodes 2, 7, 1, and 3. The sum is 2 + 7 + 1 + 3 = 13. Wait, let me recalculate: Looking at grandparent 6, its children are 7 and 8, so grandchildren are 2, 7 (left subtree) and 1, 3 (right subtree). Actually checking nodes with even grandparent 6: we get 2+7+1+3=13. But the expected is 18, so there might be another even grandparent like 8. Node 8's grandchildren would be: none from left child 1, and node 5 from right child 3. So total: 2+7+1+3+5=18.

example_2.py — Single Node

$ Input: root = [1]

› Output: 0

💡 Note: There are no grandparents in a tree with only one node, so the sum is 0.

example_3.py — No Even Grandparents

$ Input: root = [1,3,5,7,9]

› Output: 0

💡 Note: All values are odd, so there are no even-valued grandparents. The sum is 0.

Visualization

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Understanding the Visualization

Brute Force: Ask Everyone

Go to each person and ask: 'Do you have a grandparent with an even bank account?' This requires everyone to remember their family tree.

Smart Approach: Pass Information Down

Start from the eldest generation and pass down information: 'Your grandparent has account #6 (even), so you get inheritance!'

Optimal Traversal

In one family reunion, efficiently pass grandparent info to each generation. No need to ask backward, just pass forward.

Calculate Total

Sum up all the inheritance money distributed to grandchildren of wealthy (even-numbered) grandparents.

Key Takeaway

🎯 Key Insight: Instead of each grandchild searching backward for their grandparent, we pass the grandparent's information forward during our family tree traversal. This transforms an O(n²) backward-lookup problem into an elegant O(n) forward-passing solution!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each node exactly once during DFS traversal

✓ Linear Growth

Space Complexity

O(h)

Recursion stack depth equals tree height, O(log n) for balanced tree, O(n) worst case

✓ Linear Space

Constraints

The number of nodes in the tree is in the range [1, 10⁴]
1 ≤ Node.val ≤ 100
Tree nodes are connected in a valid binary tree structure

Asked in

f Facebook 15 a Amazon 12 G Google 8 ⊞ Microsoft 6

The optimal solution uses single-pass DFS with parameter passing. Instead of storing parent relationships, we pass grandparent and parent information down during traversal. At each node, we check if the grandparent exists and has an even value, then add the current node's value to our sum. This achieves O(n) time and O(h) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Find Grandparent for Each Node)	O(n²)	O(n)	For each node, search back up to find its grandparent and check if even
Single Pass DFS (Optimal)	O(n)	O(h)	Pass grandparent and parent info down during DFS traversal

Brute Force (Find Grandparent for Each Node) — Algorithm Steps

Traverse the tree and build parent-child relationships
For each node, check if it has a grandparent
If grandparent exists and has even value, add current node to sum
Return the total sum

Visualization

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Step-by-Step Walkthrough

Build Parent Map

First pass: create mapping from each node to its parent

Check Each Node

For every node, look up its parent, then grandparent

Add to Sum

If grandparent exists and is even, add current node value

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

int dfs(struct TreeNode* node, struct TreeNode* parent, struct TreeNode* grandparent) {
    if (!node) return 0;
    
    int sum = 0;
    if (grandparent && grandparent->val % 2 == 0) {
        sum += node->val;
    }
    
    sum += dfs(node->left, node, parent);
    sum += dfs(node->right, node, parent);
    
    return sum;
}

int sumEvenGrandparent(struct TreeNode* root) {
    return dfs(root, NULL, NULL);
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each node, we might need to traverse up to find grandparent

⚠ Quadratic Growth

Space Complexity

O(n)

Need to store parent relationships for all nodes

⚡ Linearithmic Space

Constraints

The number of nodes in the tree is in the range [1, 10⁴]
1 ≤ Node.val ≤ 100
Tree nodes are connected in a valid binary tree structure

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Smart Actions

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Find Grandparent for Each Node) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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