# Sum of all the child nodes with even grandparents in a Binary Tree

Given a Binary Tree, calculate the sum of nodes with even valued Grandparents.
Examples:

```Input:
22
/    \
3      8
/ \    / \
4   8  1   9
\
2
Output: 24
Explanation
The nodes 4, 8, 2, 1, 9
has even value grandparents.
Hence sum = 4 + 8 + 1 + 9 + 2 = 24.

Input:
1
/   \
2     3
/ \   / \
4   5 6   7
/
8
Output: 8
Explanation
Only 8 has 2 as a grandparent.
```

Approach: To solve the problem mentioned above, for each node that is not null, check if they have a grandparent and if their grandparent is even valued add the node’s data to the sum.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to find sum` `// of all the child nodes with` `// even grandparents in a Binary Tree`   `#include ` `using` `namespace` `std;`   `/* A binary tree node has data and` `pointers to the right and left children*/` `struct` `TreeNode {` `    ``int` `data;` `    ``TreeNode *left, *right;` `    ``TreeNode(``int` `x)` `    ``{` `        ``data = x;` `        ``left = right = NULL;` `    ``}` `};`   `// Function to calculate the sum` `void` `getSum(` `    ``TreeNode* curr, TreeNode* p,` `    ``TreeNode* gp, ``int``& sum)` `{` `    ``// Base condition` `    ``if` `(curr == NULL)` `        ``return``;`   `    ``// Check if node has a grandparent` `    ``// if it does check` `    ``// if they are even valued` `    ``if` `(gp != NULL && gp->data % 2 == 0)` `        ``sum += curr->data;`   `    ``// Recurse for left child` `    ``getSum(curr->left, curr, p, sum);`   `    ``// Recurse for right child` `    ``getSum(curr->right, curr, p, sum);` `}`   `// Driver Program` `int` `main()` `{` `    ``TreeNode* root = ``new` `TreeNode(22);`   `    ``root->left = ``new` `TreeNode(3);` `    ``root->right = ``new` `TreeNode(8);`   `    ``root->left->left = ``new` `TreeNode(4);` `    ``root->left->right = ``new` `TreeNode(8);`   `    ``root->right->left = ``new` `TreeNode(1);` `    ``root->right->right = ``new` `TreeNode(9);` `    ``root->right->right->right = ``new` `TreeNode(2);`   `    ``int` `sum = 0;` `    ``getSum(root, NULL, NULL, sum);` `    ``cout << sum << ``'\n'``;`   `    ``return` `0;` `}`

## Java

 `// Java implementation to find sum` `// of all the child nodes with` `// even grandparents in a Binary Tree` `import` `java.util.*;` `class` `GFG{`   `/* A binary tree node has data and` `pointers to the right and left children*/` `static` `class` `TreeNode ` `{` `  ``int` `data;` `  ``TreeNode left, right;` `  ``TreeNode(``int` `x)` `  ``{` `    ``data = x;` `    ``left = right = ``null``;` `  ``}` `}` `  `  `static` `int` `sum = ``0``;` `  `  `// Function to calculate the sum` `static` `void` `getSum(TreeNode curr, ` `                   ``TreeNode p,` `                   ``TreeNode gp)` `{` `  ``// Base condition` `  ``if` `(curr == ``null``)` `    ``return``;`   `  ``// Check if node has ` `  ``// a grandparent` `  ``// if it does check` `  ``// if they are even valued` `  ``if` `(gp != ``null` `&& gp.data % ``2` `== ``0``)` `    ``sum += curr.data;`   `  ``// Recurse for left child` `  ``getSum(curr.left, curr, p);`   `  ``// Recurse for right child` `  ``getSum(curr.right, curr, p);` `}`   `// Driver Program` `public` `static` `void` `main(String[] args)` `{` `  ``TreeNode root = ``new` `TreeNode(``22``);`   `  ``root.left = ``new` `TreeNode(``3``);` `  ``root.right = ``new` `TreeNode(``8``);`   `  ``root.left.left = ``new` `TreeNode(``4``);` `  ``root.left.right = ``new` `TreeNode(``8``);`   `  ``root.right.left = ``new` `TreeNode(``1``);` `  ``root.right.right = ``new` `TreeNode(``9``);` `  ``root.right.right.right = ``new` `TreeNode(``2``);`   `  ``getSum(root, ``null``, ``null``);` `  ``System.out.println(sum);` `}` `}`   `// This code is contributed by Rajput-Ji`

## C#

 `// C# implementation to find sum` `// of all the child nodes with` `// even grandparents in a Binary Tree` `using` `System;` `class` `GFG{`   `/* A binary tree node ` `has data and pointers to ` `the right and left children*/` `class` `TreeNode ` `{` `  ``public` `int` `data;` `  ``public` `TreeNode left, right;` `  ``public` `TreeNode(``int` `x)` `  ``{` `    ``data = x;` `    ``left = right = ``null``;` `  ``}` `}` `  `  `static` `int` `sum = 0;` `  `  `// Function to calculate the sum` `static` `void` `getSum(TreeNode curr, ` `                   ``TreeNode p,` `                   ``TreeNode gp)` `{` `  ``// Base condition` `  ``if` `(curr == ``null``)` `    ``return``;`   `  ``// Check if node has ` `  ``// a grandparent` `  ``// if it does check` `  ``// if they are even valued` `  ``if` `(gp != ``null` `&& gp.data % 2 == 0)` `    ``sum += curr.data;`   `  ``// Recurse for left child` `  ``getSum(curr.left, curr, p);`   `  ``// Recurse for right child` `  ``getSum(curr.right, curr, p);` `}`   `// Driver Program` `public` `static` `void` `Main(String[] args)` `{` `  ``TreeNode root = ``new` `TreeNode(22);`   `  ``root.left = ``new` `TreeNode(3);` `  ``root.right = ``new` `TreeNode(8);`   `  ``root.left.left = ``new` `TreeNode(4);` `  ``root.left.right = ``new` `TreeNode(8);`   `  ``root.right.left = ``new` `TreeNode(1);` `  ``root.right.right = ``new` `TreeNode(9);` `  ``root.right.right.right = ``new` `TreeNode(2);`   `  ``getSum(root, ``null``, ``null``);` `  ``Console.WriteLine(sum);` `}` `}`   `// This code is contributed by Princi Singh`

Output:

```24

```

Time Complexity: O(N)
Space Complexity: O(H), Used by recursion stack where H = height of the tree.

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Improved By : Rajput-Ji, princi singh