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Traversal in Doubly Linked List
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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




Next Article
Traversal in Doubly Linked List

K
Kanishk_Verma
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Article Tags :
  • DSA
  • Linked List
  • doubly linked list
Practice Tags :
  • Linked List

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