Find the second last node of a linked list in single traversal

Given a linked list. The task is to find the second last node of the linked list using a single traversal only.

Examples:

Input : List = 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output : 4

Input : List = 2 -> 4 -> 6 -> 8 -> 33 -> 67 -> NULL
Output : 33

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to traverse the linked list following the below approach:

If the list is empty or contains less than 2 elements, return false.
Otherwise check if the current node is the second last node of the linked list or not. That is, if (current_node->next-next == NULL ) then the current node is the second last node.
If the current node is the second last node, print the node otherwise move to the next node.
Repeat the above two steps until the second last node is reached.

Below is the implementation of the above approach:

C++

// C++ program to find the second last node 
// of a linked list in single traversal 
  
#include <iostream> 
using namespace std; 
  
// Link list node 
struct Node { 
    int data; 
    struct Node* next; 
}; 
  
// Function to find the second last 
// node of the linked list 
int findSecondLastNode(struct Node* head) 
{ 
    struct Node* temp = head; 
  
    // If the list is empty or contains less 
    // than 2 nodes 
    if (temp == NULL || temp->next == NULL) 
        return -1; 
  
    // Traverse the linked list 
    while (temp != NULL) { 
  
        // Check if the current node  is the 
        // second last node or not 
        if (temp->next->next == NULL) 
            return temp->data; 
  
        // If not then move to the next node 
        temp = temp->next; 
    } 
} 
  
// Function to push node at head 
void push(struct Node** head_ref, int new_data) 
{ 
    Node* new_node = new Node; 
    new_node->data = new_data; 
    new_node->next = (*head_ref); 
    (*head_ref) = new_node; 
} 
  
// Driver code 
int main() 
{ 
    /* Start with the empty list */
    struct Node* head = NULL; 
  
    /* Use push() function to construct  
    the below list 8 -> 23 -> 11 -> 29 -> 12 */
    push(&head, 12); 
    push(&head, 29); 
    push(&head, 11); 
    push(&head, 23); 
    push(&head, 8); 
  
    cout << findSecondLastNode(head); 
  
    return 0; 
} 

Java

// Java program to find the second last node  
// of a linked list in single traversal 
  
// Linked list node 
class Node 
{ 
    int data; 
    Node next; 
    Node(int d) 
    { 
        this.data = d; 
        this.next = null; 
    } 
} 
class LinkedList 
{ 
    Node start; 
    LinkedList() 
    { 
        start = null; 
    } 
      
    // Function to push node at head 
    public void push(int data) 
    {  
        if(this.start == null) 
        { 
        Node temp = new Node(data); 
        this.start = temp; 
        } 
        else
        { 
            Node temp = new Node(data); 
            temp.next = this.start; 
            this.start = temp; 
        } 
    } 
      
    // method to find the second last  
    // node of the linked list  
    public int findSecondLastNode(Node ptr) 
    { 
        Node temp = ptr; 
          
        // If the list is empty or contains less  
        // than 2 nodes 
        if(temp == null || temp.next == null) 
            return -1; 
      
            // This loop stops at second last node 
        while(temp.next.next != null) 
        { 
            temp = temp.next; 
        } 
        return temp.data; 
    } 
      
    // Driver code 
    public static void main(String[] args) 
    { 
        LinkedList ll = new LinkedList(); 
          
        /* Use push() function to construct  
        the below list 8 -> 23 -> 11 -> 29 -> 12 */
        ll.push(12); 
        ll.push(29); 
        ll.push(11); 
        ll.push(23); 
        ll.push(8); 
        System.out.println(ll.findSecondLastNode(ll.start)); 
    } 
} 
  
// This code is Contributed by Adarsh_Verma 

Python3

# Python3 program to find the second last node 
# of a linked list in single traversal 
import math 
  
# Link list node 
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.next = None
  
# Function to find the second last 
# node of the linked list 
def findSecondLastNode(head): 
    temp = head 
  
    # If the list is empty or  
    # contains less than 2 nodes 
    if (temp == None or temp.next == None): 
        return -1
  
    # Traverse the linked list 
    while (temp != None): 
  
        # Check if the current node is the 
        # second last node or not 
        if (temp.next.next == None): 
            return temp.data 
              
        # If not then move to the next node 
        temp = temp.next
  
# Function to push node at head 
def push(head, new_data): 
    new_node = Node(new_data) 
    #new_node.data = new_data 
    new_node.next = head 
    head = new_node 
    return head 
  
# Driver code 
if __name__ == '__main__': 
      
    # Start with the empty list 
    head = None
  
    # Use push() function to construct 
    # the below list 8 . 23 . 11 . 29 . 12 
    head = push(head, 12) 
    head = push(head, 29) 
    head = push(head, 11) 
    head = push(head, 23) 
    head = push(head, 8) 
  
    print(findSecondLastNode(head)) 
  
# This code is contributed by Srathore 

C#

// C# program to find the second last node  
// of a linked list in single traversal 
using System; 
  
// Linked list node 
public class Node 
{ 
    public int data; 
    public Node next; 
    public Node(int d) 
    { 
        this.data = d; 
        this.next = null; 
    } 
} 
public class LinkedList 
{ 
    Node start; 
    LinkedList() 
    { 
        start = null; 
    } 
      
    // Function to push node at head 
    public void push(int data) 
    {  
        if(this.start == null) 
        { 
        Node temp = new Node(data); 
        this.start = temp; 
        } 
        else
        { 
            Node temp = new Node(data); 
            temp.next = this.start; 
            this.start = temp; 
        } 
    } 
      
    // method to find the second last  
    // node of the linked list  
    public int findSecondLastNode(Node ptr) 
    { 
        Node temp = ptr; 
          
        // If the list is empty or contains less  
        // than 2 nodes 
        if(temp == null || temp.next == null) 
            return -1; 
      
            // This loop stops at second last node 
        while(temp.next.next != null) 
        { 
            temp = temp.next; 
        } 
        return temp.data; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        LinkedList ll = new LinkedList(); 
          
        /* Use push() function to construct  
        the below list 8 -> 23 -> 11 -> 29 -> 12 */
        ll.push(12); 
        ll.push(29); 
        ll.push(11); 
        ll.push(23); 
        ll.push(8); 
        Console.WriteLine(ll.findSecondLastNode(ll.start)); 
    } 
} 
  
// This code is contributed by PrinciRaj1992 

Output:

Time complexity : O(n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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