Skip to content
Related Articles

Related Articles

Sum of all the parent nodes having child node x
  • Difficulty Level : Easy
  • Last Updated : 06 Nov, 2020
GeeksforGeeks - Summer Carnival Banner

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).

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes
arrow_drop_up
Next

Find sum of all left leaves in a given Binary Tree
Recommended Articles
Page :
Sum of all the child nodes with even parent values in a Binary Tree
15, Jan 20
Sum of all parent-child differences in a Binary Tree
29, Apr 19
Count all Grandparent-Parent-Child Triplets in a binary tree whose sum is greater than X
18, Dec 18
Ways to color a skewed tree such that parent and child have different colors
10, Nov 17
Print the nodes having exactly one child in a Binary tree
04, May 20
Sum of all the child nodes with even grandparents in a Binary Tree
16, May 20
Count of nodes in a Binary Tree whose child is its prime factors
09, Apr 20
Find parent of each node in a tree for multiple queries
20, Nov 19
Find parent of given node in a Binary Tree with given postorder traversal
09, Jun 20
Find the parent of a node in the given binary tree
11, Nov 19
Maximum parent children sum in Binary tree
04, Sep 18
Sum of all odd nodes in the path connecting two given nodes
28, Nov 18
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
19, Mar 21
Find a node such that all paths from that node to leaf nodes are of the same color
20, Jan 21
Sum of nodes having sum of subtrees of opposite parities
23, Nov 20
Node having maximum number of nodes less than its value in its subtree
10, Sep 20
Find Height of Binary Tree represented by Parent array
05, Oct 14
Construct Binary Tree from given Parent Array representation
16, Oct 15
Lowest Common Ancestor in a Binary Tree | Set 2 (Using Parent Pointer)
07, Mar 16
Construct Binary Tree from given Parent Array representation | Iterative Approach
20, Mar 19
Height of a generic tree from parent array
23, Jan 18
Height of n-ary tree if parent array is given
30, Aug 17
Find right sibling of a binary tree with parent pointers
20, Aug 17
Lowest Common Ancestor in Parent Array Representation
11, Sep 18
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Easy





Improved By :
Article Tags :
Practice Tags :
Report Issue

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

Lightbox