Open In App

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy

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

Implementation:






// 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 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 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


















                        






                        























                                    



                                    




                                    




                                    


                                




















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


















                        






                        























                                    



                                    




                                    




                                    


                                



















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

	

        Article Tags : 
        

            DSA
Tree        


	


      

      

          
Recommended Articles

      

      

          
          
1. Sum of all the child nodes with even parent values in a Binary Tree

          
2. Sum of nodes in a binary tree having only the left child nodes

          
3. Count all Grandparent-Parent-Child Triplets in a binary tree whose sum is greater than X

          
4. Sum of all parent-child differences in a Binary Tree

          
5. Count nodes with sum of path made by only left child nodes at least K

          
6. Identify all Grand-Parent Nodes of each Node in a Map

          
7. Ways to color a skewed tree such that parent and child have different colors

          
8. Using fork() to produce 1 parent and its 3 child processes

          
9. Print the nodes having exactly one child in a Binary tree

          
10. Sum of all the child nodes with even grandparents in a Binary Tree

          
11. Find a set of at most N/2 nodes from a Graph such that all remaining nodes are directly connected to one of the chosen nodes

          
12. Count of nodes in a given N-ary tree having distance to all leaf nodes equal in their subtree

          
13. Find node having maximum number of common nodes with a given node K

          
14. Count of nodes in a binary tree having their nodes in range [L, R]

          
15. Count of nodes in a Binary Tree whose child is its prime factors

          
16. Sum of all odd nodes in the path connecting two given nodes

          
17. Construct a Tree whose sum of nodes of all the root to leaf path is not divisible by the count of nodes in that path

          
18. Sum of nodes having sum of subtrees of opposite parities

          
19. Find a node such that all paths from that node to leaf nodes are of the same color

          
20. Kth largest node among all directly connected nodes to the given node in an undirected graph

          
21. Find parent of each node in a tree for multiple queries

          
22. Find the parent of a node in the given binary tree

          
23. Find parent of given node in a Binary Tree with given postorder traversal

          
24. Count of graphs formed by changing color of any red-colored node with black parent to black

          
25. Find the parent node of maximum product Siblings in given Binary Tree