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Program to find size of Doubly Linked List

Last Updated : 04 Mar, 2025
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Given a doubly linked list, The task is to find the size of the given doubly linked list.

Example:

Input : 1<->2<->3<->4
output : 4
Explanation: The size is 4 as there are 4 nodes in the doubly linked list.

Input : 1<->2
output : 2

Approach - Using While Loop – O(n) Time and O(1) Space

The idea is to traverse the doubly linked list starting from the head node. Increment the size variable until we reach the end.

C++ 
#include <bits/stdc++.h>
using namespace std;

class Node {
  public:
    int data;
    Node *next;
    Node *prev;
    Node(int val) {
        data = val;
        next = nullptr;
        prev = nullptr;
    }
};

// This function returns size of linked list
int findSize(Node *curr) {
    int size = 0;
    while (curr != NULL) {
        size++;
        curr = curr->next;
    }
    return size;
}

int main() {
  
    // Create a hard-coded doubly linked list:
    // 1 <-> 2 <-> 3 <-> 4
    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->prev = head;
    head->next->next = new Node(3);
    head->next->next->prev = head->next;
    head->next->next->next = new Node(4);
    head->next->next->next->prev = head->next->next;

    cout << findSize(head);
    return 0;
}
C 
#include <stdio.h>

struct Node {
  int data;
  struct Node* prev;
  struct Node* next;
};


int findSize(struct Node* curr) {
  int size = 0;
  while (curr != NULL) {
    size++;
    curr = curr->next;
  }
  return size;
}

struct Node *createNode(int new_data) { 
    struct Node *new_node = 
      (struct Node *)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = NULL;
    new_node->prev = NULL;
    return new_node;
}

int main() {
  
	// Create a hard-coded doubly linked list:
    // 1 <-> 2 <-> 3 <-> 4
    struct Node *head = createNode(1);
    head->next = createNode(2);
    head->next->prev = head;
    head->next->next = createNode(3);
    head->next->next->prev = head->next;
    head->next->next->next = createNode(4);
    head->next->next->next->prev = head->next->next;
  	printf("%d\n", findSize(head));

  return 0;
}
Java 
class Node {
    int data;
    Node next;
    Node prev;
    
    Node(int val) {
        data = val;
        next = null;
        prev = null;
    }
}

public class GfG {
    // This function returns the size of 
  	// the linked list
    static int findSize(Node curr) {
        int size = 0;
        while (curr != null) {
            size++;
            curr = curr.next;
        }
        return size;
    }

    public static void main(String[] args) {
        // Create a hard-coded doubly linked list:
        // 1 <-> 2 <-> 3 <-> 4
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.prev = head;
        head.next.next = new Node(3);
        head.next.next.prev = head.next;
        head.next.next.next = new Node(4);
        head.next.next.next.prev = head.next.next;

        System.out.println(findSize(head)); 
    }
}
Python 
class Node:
    def __init__(self, val):
        self.data = val
        self.next = None
        self.prev = None

# This function returns the size of
# the linked list
def findSize(curr):
    size = 0
    while curr:
        size += 1
        curr = curr.next
    return size

if __name__ == "__main__":
  
    # Create a hard-coded doubly linked list:
    # 1 <-> 2 <-> 3 <-> 4
    head = Node(1)
    head.next = Node(2)
    head.next.prev = head
    head.next.next = Node(3)
    head.next.next.prev = head.next
    head.next.next.next = Node(4)
    head.next.next.next.prev = head.next.next

    print(findSize(head))
C# 
using System;

class Node {
    public int data;
    public Node next;
    public Node prev;
   
    public Node(int val) {
        data = val;
        next = null;
        prev = null;
    }
}

class GfG {
    
    // This function returns the size of 
  	// the linked list
    static int findSize(Node curr) {
        int size = 0;
        while (curr != null) {
            size++;
            curr = curr.next;
        }
        return size;
    }

    static void Main() {
      
        // Create a hard-coded doubly linked list:
        // 1 <-> 2 <-> 3 <-> 4
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.prev = head;
        head.next.next = new Node(3);
        head.next.next.prev = head.next;
        head.next.next.next = new Node(4);
        head.next.next.next.prev = head.next.next;

        Console.WriteLine(findSize(head));
    }
}
JavaScript 
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
        this.prev = null;
    }
}

// This function returns the size
// of the linked list
function findSize(curr) {
    let size = 0;
    while (curr !== null) {
        size++;
        curr = curr.next;
    }
    return size;
}

// Create a hard-coded doubly linked list:
// 1 <-> 2 <-> 3 <-> 4
let head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(4);
head.next.next.next.prev = head.next.next;

console.log(findSize(head));

Output
4

Approach - Using Recursion - O(n) Time and O(n) Space

The idea is to use a recursive function that takes a node and check the next pointer of the current node if the next pointer is NULL then return 0 else if the next pointer is not NULL then return 1 + the value returned by the recursive function for the next node.

C++ 
#include <bits/stdc++.h>
using namespace std;

// Definition for doubly linked list node
struct Node {
    int data;
    Node* next;
    Node* prev;
    
    Node(int val) : data(val), next(NULL), prev(NULL) {}
};

// Function to calculate the size of the doubly linked list using recursion
int findSize(Node* head) {
    if (head == NULL)
        return 0;
    return 1 + findSize(head->next); // Recursive call
}

int main() {
    // Creating a doubly linked list
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->prev = head;
    head->next->next = new Node(3);
    head->next->next->prev = head->next;
    head->next->next->next = new Node(3);
    head->next->next->next->prev = head->next->next; 
    
    cout << findSize(head) << endl;
    return 0;
}
C 
#include <stdio.h>
#include <stdlib.h>

// Definition for doubly linked list node
struct Node {
    int data;
    struct Node* next;
    struct Node* prev;
};

// Function to calculate the size of the doubly linked list using recursion
int findSize(struct Node* head) {
    if (head == NULL)
        return 0;
    return 1 + findSize(head->next); // Recursive call
}

// Function to create a new node
struct Node* newNode(int data) {
    struct Node* node = (struct Node*)malloc(sizeof(struct Node));
    node->data = data;
    node->next = NULL;
    node->prev = NULL;
    return node;
}

int main() {
    // Creating a doubly linked list
    struct Node* head = newNode(1);
    head->next = newNode(2);
    head->next->prev = head;
    head->next->next = newNode(3);
    head->next->next->prev = head->next;
    head->next->next->next = newNode(4);
    head->next->next->next->prev = head->next->next;
    

    printf("%d\n", findSize(head));
    return 0;
}
Java 
class GfG {

    // Definition for doubly linked list node
    static class Node {
        int data;
        Node next, prev;

        Node(int data) {
            this.data = data;
            next = prev = null;
        }
    }

    // Function to calculate the size of the doubly linked list using recursion
    public static int findSize(Node head) {
        if (head == null) {
            return 0;
        }
        return 1 + findSize(head.next); // Recursive call
    }

    public static void main(String[] args) {
        // Creating a doubly linked list
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.prev = head;
        head.next.next = new Node(3);
        head.next.next.prev = head.next;
        head.next.next.next = new Node(4);
        head.next.next.next.prev = head.next.next;

        System.out.println(findSize(head));
    }
}
Python 
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
        self.prev = None

def findSize(head):
    if head is None:
        return 0
    return 1 + findSize(head.next)  # Recursive call

# Creating a doubly linked list
head = Node(1)
head.next = Node(2)
head.next.prev = head
head.next.next = Node(3)
head.next.next.prev = head.next
head.next.next.next = Node(4)
head.next.next.next.prev = head.next.next;


print(findSize(head))
C# 
using System;

class GfG
{
    // Definition for doubly linked list node
    public class Node
    {
        public int data;
        public Node next;
        public Node prev;

        public Node(int data)
        {
            this.data = data;
            this.next = null;
            this.prev = null;
        }
    }

    // Function to calculate the size of the doubly linked list using recursion
    public static int findSize(Node head)
    {
        if (head == null)
        {
            return 0;
        }
        return 1 + findSize(head.next); // Recursive call
    }

    public static void Main(string[] args)
    {
        // Creating a doubly linked list
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.prev = head;
        head.next.next = new Node(3);
        head.next.next.prev = head.next;
        head.next.next.next = new Node(4);
        head.next.next.next.prev = head.next.next;
        
        // Printing the size of the doubly linked list
        Console.WriteLine(findSize(head));
    }
}
JavaScript 
class Node {
    constructor(data) {
        this.data = data;
        this.next = null;
        this.prev = null;
    }
}

function findSize(head) {
    if (head === null) {
        return 0;
    }
    return 1 + findSize(head.next);  // Recursive call
}

// Creating a doubly linked list
const head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(4);
head.next.next.next.prev = head.next.next;

console.log(findSize(head));

Output
Size of the doubly linked list: 4



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