# Sum of all the parent nodes having child node x

• Difficulty Level : Easy
• Last Updated : 16 Aug, 2022

Given a binary tree containing n nodes. The problem is to find the sum of all the parent node’s which have a child node with value x.

Examples:

```Input : Binary tree with x = 2:
4
/ \
2   5
/ \ / \
7  2 2  3
Output : 11

4
/ \
2   5
/ \ / \
7  2 2  3

The highlighted nodes (4, 2, 5) above
are the nodes having 2 as a child node.```

Algorithm:

```sumOfParentOfX(root,sum,x)
if root == NULL
return

if (root->left && root->left->data == x) ||
(root->right && root->right->data == x)
sum += root->data

sumOfParentOfX(root->left, sum, x)
sumOfParentOfX(root->right, sum, x)

sumOfParentOfXUtil(root,x)
Declare sum = 0
sumOfParentOfX(root, sum, x)
return sum```

Implementation:

## C++

 `// C++ implementation to find the sum of all``// the parent nodes having child node x``#include ` `using` `namespace` `std;` `// Node of a binary tree``struct` `Node``{``    ``int` `data;``    ``Node *left, *right;``};` `// function to get a new node``Node* getNode(``int` `data)``{``    ``// allocate memory for the node``    ``Node *newNode =``        ``(Node*)``malloc``(``sizeof``(Node));``    ` `    ``// put in the data   ``    ``newNode->data = data;``    ``newNode->left = newNode->right = NULL;``    ``return` `newNode;   ``}` `// function to find the sum of all the``// parent nodes having child node x``void` `sumOfParentOfX(Node* root, ``int``& sum, ``int` `x)``{``    ``// if root == NULL``    ``if` `(!root)``        ``return``;``    ` `    ``// if left or right child of root is 'x', then``    ``// add the root's data to 'sum'``    ``if` `((root->left && root->left->data == x) ||``        ``(root->right && root->right->data == x))``        ``sum += root->data;``    ` `    ``// recursively find the required parent nodes``    ``// in the left and right subtree   ``    ``sumOfParentOfX(root->left, sum, x);``    ``sumOfParentOfX(root->right, sum, x);``    ` `}` `// utility function to find the sum of all``// the parent nodes having child node x``int` `sumOfParentOfXUtil(Node* root, ``int` `x)``{``    ``int` `sum = 0;``    ``sumOfParentOfX(root, sum, x);``    ` `    ``// required sum of parent nodes``    ``return` `sum;``}` `// Driver program to test above``int` `main()``{``    ``// binary tree formation``    ``Node *root = getNode(4);           ``/*        4        */``    ``root->left = getNode(2);           ``/*       / \       */``    ``root->right = getNode(5);          ``/*      2   5      */``    ``root->left->left = getNode(7);     ``/*     / \ / \     */``    ``root->left->right = getNode(2);    ``/*    7  2 2  3    */``    ``root->right->left = getNode(2);``    ``root->right->right = getNode(3);``    ` `    ``int` `x = 2;``    ` `    ``cout << ``"Sum = "``         ``<< sumOfParentOfXUtil(root, x);``         ` `    ``return` `0;   ``}`

## Java

 `// Java implementation to find``// the sum of all the parent``// nodes having child node x``class` `GFG``{``// sum``static` `int` `sum = ``0``;``    ` `    ` `// Node of a binary tree``static` `class` `Node``{``    ``int` `data;``    ``Node left, right;``};` `// function to get a new node``static` `Node getNode(``int` `data)``{``    ``// allocate memory for the node``    ``Node newNode = ``new` `Node();``    ` `    ``// put in the data    ``    ``newNode.data = data;``    ``newNode.left = newNode.right = ``null``;``    ``return` `newNode;    ``}` `// function to find the sum of all the``// parent nodes having child node x``static` `void` `sumOfParentOfX(Node root, ``int` `x)``{``    ``// if root == NULL``    ``if` `(root == ``null``)``        ``return``;``    ` `    ``// if left or right child``    ``// of root is 'x', then``    ``// add the root's data to 'sum'``    ``if` `((root.left != ``null` `&& root.left.data == x) ||``        ``(root.right != ``null` `&& root.right.data == x))``        ``sum += root.data;``    ` `    ``// recursively find the required``    ``// parent nodes in the left and``    ``// right subtree    ``    ``sumOfParentOfX(root.left, x);``    ``sumOfParentOfX(root.right, x);``    ` `}` `// utility function to find the``// sum of all the parent nodes``// having child node x``static` `int` `sumOfParentOfXUtil(Node root,   ``                              ``int` `x)``{``    ``sum = ``0``;``    ``sumOfParentOfX(root, x);``    ` `    ``// required sum of parent nodes``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``// binary tree formation``    ``Node root = getNode(``4``);         ``//     4    ``    ``root.left = getNode(``2``);         ``//     / \    ``    ``root.right = getNode(``5``);         ``//     2 5    ``    ``root.left.left = getNode(``7``);     ``//     / \ / \    ``    ``root.left.right = getNode(``2``); ``// 7 2 2 3``    ``root.right.left = getNode(``2``);``    ``root.right.right = getNode(``3``);``    ` `    ``int` `x = ``2``;``    ` `    ``System.out.println( ``"Sum = "` `+``           ``sumOfParentOfXUtil(root, x));``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation to find the Sum of``# all the parent nodes having child node x` `# function to get a new node``class` `getNode:``    ``def` `__init__(``self``, data):``        ` `        ``# put in the data    ``        ``self``.data ``=` `data``        ``self``.left ``=` `self``.right ``=` `None` `# function to find the Sum of all the``# parent nodes having child node x``def` `SumOfParentOfX(root, ``Sum``, x):``    ` `    ``# if root == None``    ``if` `(``not` `root):``        ``return``    ` `    ``# if left or right child of root is 'x',``    ``# then add the root's data to 'Sum'``    ``if` `((root.left ``and` `root.left.data ``=``=` `x) ``or``        ``(root.right ``and` `root.right.data ``=``=` `x)):``        ``Sum``[``0``] ``+``=` `root.data``    ` `    ``# recursively find the required parent``    ``# nodes in the left and right subtree    ``    ``SumOfParentOfX(root.left, ``Sum``, x)``    ``SumOfParentOfX(root.right, ``Sum``, x)` `# utility function to find the Sum of all``# the parent nodes having child node x``def` `SumOfParentOfXUtil(root, x):``    ``Sum` `=` `[``0``]``    ``SumOfParentOfX(root, ``Sum``, x)``    ` `    ``# required Sum of parent nodes``    ``return` `Sum``[``0``]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# binary tree formation``    ``root ``=` `getNode(``4``)         ``#     4    ``    ``root.left ``=` `getNode(``2``)         ``#     / \    ``    ``root.right ``=` `getNode(``5``)         ``#     2 5    ``    ``root.left.left ``=` `getNode(``7``)     ``#     / \ / \    ``    ``root.left.right ``=` `getNode(``2``) ``# 7 2 2 3``    ``root.right.left ``=` `getNode(``2``)``    ``root.right.right ``=` `getNode(``3``)``    ` `    ``x ``=` `2``    ` `    ``print``(``"Sum = "``, SumOfParentOfXUtil(root, x))``    ` `# This code is contributed by PranchalK`

## C#

 `using` `System;` `// C# implementation to find``// the sum of all the parent``// nodes having child node x``public` `class` `GFG``{``// sum``public` `static` `int` `sum = 0;`  `// Node of a binary tree``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;``}` `// function to get a new node``public` `static` `Node getNode(``int` `data)``{``    ``// allocate memory for the node``    ``Node newNode = ``new` `Node();` `    ``// put in the data    ``    ``newNode.data = data;``    ``newNode.left = newNode.right = ``null``;``    ``return` `newNode;``}` `// function to find the sum of all the``// parent nodes having child node x``public` `static` `void` `sumOfParentOfX(Node root, ``int` `x)``{``    ``// if root == NULL``    ``if` `(root == ``null``)``    ``{``        ``return``;``    ``}` `    ``// if left or right child``    ``// of root is 'x', then``    ``// add the root's data to 'sum'``    ``if` `((root.left != ``null` `&& root.left.data == x) ||``       ``(root.right != ``null` `&& root.right.data == x))``    ``{``        ``sum += root.data;``    ``}` `    ``// recursively find the required``    ``// parent nodes in the left and``    ``// right subtree    ``    ``sumOfParentOfX(root.left, x);``    ``sumOfParentOfX(root.right, x);` `}` `// utility function to find the``// sum of all the parent nodes``// having child node x``public` `static` `int` `sumOfParentOfXUtil(Node root, ``int` `x)``{``    ``sum = 0;``    ``sumOfParentOfX(root, x);` `    ``// required sum of parent nodes``    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``// binary tree formation``    ``Node root = getNode(4); ``//     4``    ``root.left = getNode(2); ``//     / \``    ``root.right = getNode(5); ``//     2 5``    ``root.left.left = getNode(7); ``//     / \ / \``    ``root.left.right = getNode(2); ``// 7 2 2 3``    ``root.right.left = getNode(2);``    ``root.right.right = getNode(3);` `    ``int` `x = 2;` `    ``Console.WriteLine(``"Sum = "` `+ sumOfParentOfXUtil(root, x));``}``}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`Sum = 11`

Time Complexity: O(n).

Auxiliary space: O(n) for call stack as it is using recursion

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