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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.



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).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

tags:

Tree Tree

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