Sum of the nodes of a Singly Linked List

Given a singly linked list. The task is to find the sum of nodes of the given linked list.

Task is to do A + B + C + D.


Examples:

Input: 7->6->8->4->1
Output: 26
Sum of nodes:
7 + 6 + 8 + 4 + 1 = 26

Input: 1->7->3->9->11->5
Output: 36


Recursive Solution:

  • Call a function by passing the head and variable to store the sum.
    • Then recursively call the function by passing the next of current node and sum variable.
      • Add the value of the current node to the sum.

Below is the implementation of above approach:

C++

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// C++ implementation to find the sum of
// nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = new Node;
  
    /* put in the data */
    new_node->data = new_data;
  
    /* link the old list to the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
// function to recursively find the sum of
// nodes of the given linked list
void sumOfNodes(struct Node* head, int* sum)
{
    // if head = NULL
    if (!head)
        return;
  
    // recursively traverse the remaining nodes
    sumOfNodes(head->next, sum);
  
    // accumulate sum
    *sum = *sum + head->data;
}
  
// utility function to find the sum of  nodes
int sumOfNodesUtil(struct Node* head)
{
  
    int sum = 0;
  
    // find the sum of  nodes
    sumOfNodes(head, &sum);
  
    // required sum
    return sum;
}
  
// Driver program to test above
int main()
{
    struct Node* head = NULL;
  
    // create linked list 7->6->8->4->1
    push(&head, 7);
    push(&head, 6);
    push(&head, 8);
    push(&head, 4);
    push(&head, 1);
  
    cout << "Sum of nodes = "
         << sumOfNodesUtil(head);
    return 0;
}

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Java

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// Java implementation to find the sum of 
// nodes of the Linked List 
class GFG 
{
      
static int sum=0;
      
// A Linked list node /
static class Node
    int data; 
    Node next; 
  
// function to insert a node at the 
// beginning of the linked list 
static Node push( Node head_ref, int new_data) 
    // allocate node /
    Node new_node = new Node(); 
  
    // put in the data /
    new_node.data = new_data; 
  
    // link the old list to the new node /
    new_node.next = (head_ref); 
  
    // move the head to point to the new node /
    (head_ref) = new_node; 
    return head_ref;
  
// function to recursively find the sum of 
// nodes of the given linked list 
static void sumOfNodes( Node head) 
    // if head = null 
    if (head == null
        return
  
    // recursively traverse the remaining nodes 
    sumOfNodes(head.next); 
  
    // accumulate sum 
    sum = sum + head.data; 
  
// utility function to find the sum of nodes 
static int sumOfNodesUtil( Node head) 
  
    sum = 0
  
    // find the sum of nodes 
    sumOfNodes(head); 
  
    // required sum 
    return sum; 
  
// Driver program to test above 
public static void main(String args[])
    Node head = null
  
    // create linked list 7.6.8.4.1 
    head = push(head, 7); 
    head = push(head, 6); 
    head = push(head, 8); 
    head = push(head, 4); 
    head = push(head, 1); 
  
    System.out.println( "Sum of nodes = "
        + sumOfNodesUtil(head)); 
}
  
// This code is contributed by Arnab Kundu

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Python3

# Python3 implementation to find the sum of
# nodes of the Linked List
import math

# class for a Sum
class Sum:
tsum = None

# A Linked list node
class Node:
def __init__(self,data):
self.data = data
self.next = None

# function to insert a node at the
# beginning of the linked list
def push(head, data):
if not head:
return Node(data)

# put in the data
# and allocate node
new_node = Node(data)

# link the old list to the new node
new_node.next = head

# move the head to point
# to the new node
head = new_node
return head

# function to recursively find
# the sum of nodes of the given
# linked list
def sumOfNode(head, S):

# if head = NULL
if not head:
return

# recursively traverse the
# remaining nodes
sumOfNode(head.next, S)

# accumulate sum
S.tsum += head.data

# utility function to find
# the sum of nodes
def sumOfNodesUtil(head):
S = Sum()
S.tsum = 0

# find the sum of nodes
sumOfNode(head, S)

# required sum
return S.tsum

# Driver Code
if __name__==’__main__’:
head = None

# create linked list 7->6->8->4->1
head = push(head, 7)
head = push(head, 6)
head = push(head, 8)
head = push(head, 4)
head = push(head, 1)
print(“Sum of Nodes = {}” .
format(sumOfNodesUtil(head)))

# This code is contributed
# by Vikash Kumar 37

Output:

Sum of nodes = 26

Time Complexity: O(N) , N is the number of nodes in a linked list.
Auxiliary Space: O(N), only if the stack size is considered during recursive calls.

Iterative Solution:

  1. Initialise a pointer ptr with the head of the linked list and a sum variable with 0.
  2. Start traversing the linked list using a loop until all the nodes get traversed.
    • Add the value of current node to the sum i.e. sum += ptr -> data .
    • Increment the pointer to the next node of linked list i.e. ptr = ptr ->next .
  3. Return the sum.

Below is the implementation of above approach:

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code

// C++ implementation to find the sum of
// nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
  
/* A Linked list node */
struct Node {
    int data;
    struct Node* next;
};
  
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = new Node;
  
    /* put in the data */
    new_node->data = new_data;
  
    /* link the old list to the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
// function to find the sum of
// nodes of the given linked list
int sumOfNodes(struct Node* head)
{
    struct Node* ptr = head;
    int sum = 0;
    while (ptr != NULL) {
  
        sum += ptr->data;
        ptr = ptr->next;
    }
  
    return sum;
}
  
// Driver program to test above
int main()
{
    struct Node* head = NULL;
  
    // create linked list 7->6->8->4->1
    push(&head, 7);
    push(&head, 6);
    push(&head, 8);
    push(&head, 4);
    push(&head, 1);
  
    cout << "Sum of nodes = "
         << sumOfNodes(head);
    return 0;
}

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Output:

Sum of nodes = 26

Time Complexity: O(N) , N is the number of nodes in a linked list.
Auxiliary Space: O(1)



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Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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