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:

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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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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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edit
close

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link
brightness_4
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)



My Personal Notes arrow_drop_up

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

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