Sum of all the parent nodes having child node x

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

C++

// C++ implementation to find the sum of all  
// the parent nodes having child node x 
#include <bits/stdc++.h> 
  
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 

Output:

Sum = 11

Time Complexity: O(n).

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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