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

Input:22 / \ 3 8 / \ / \ 4 8 1 9 \ 2Output:24ExplanationThe 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 / 8Output:8ExplanationOnly 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 <bits/stdc++.h>` `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;` `}` |

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## 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` |

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## 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` |

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**Output:**

24

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

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