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Sum of all the parent nodes having child node x

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

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

 

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.
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