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Sum of all the parent nodes having child node x

  • Difficulty Level : Easy
  • Last Updated : 30 Jun, 2021

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.

 

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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
Javascript



<script>
 
    // JavaScript implementation to find
    // the sum of all the parent
    // nodes having child node x
     
    // sum
    let sum = 0;
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // function to get a new node
    function getNode(data)
    {
        // allocate memory for the node
        let newNode = new Node(data);
        return newNode;    
    }
 
    // function to find the sum of all the
    // parent nodes having child node x
    function sumOfParentOfX(root, 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
    function sumOfParentOfXUtil(root, x)
    {
        sum = 0;
        sumOfParentOfX(root, x);
 
        // required sum of parent nodes
        return sum;
    }
     
    // binary tree formation
    let 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);
       
    let x = 2;
       
    document.write( "Sum = " +
           sumOfParentOfXUtil(root, x));
 
</script>
Output:  
Sum = 11
Time Complexity: O(n). 
 
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 



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