# Function to check if a singly linked list is palindrome

Last Updated : 02 Apr, 2024
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Given a singly linked list of characters, write a function that returns true if the given list is a palindrome, else false.

Examples:

Input: R->A->D->A->R->NULL
Output: Yes

Input: C->O->D->E->NULL
Output: No

## Check if a Singly Linked List is Palindrome using Stack:

The idea is to use a stack and push all the nodes into the stack, then again iterate over the linked list to validate if the linked list is palindrome or not.

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Below image is a dry run of the above approach:Â

Follow the steps below to solve the problem:

• A simple solution is to use a stack of list nodes. This mainly involves three steps.
• Traverse the given list from head to tail and push every visited node to stack.
• Traverse the list again. For every visited node, pop a node from the stack and compare data of popped node with the currently visited node.
• If all nodes matched, then return true, else false.

Below is the implementation of the above approach :Â

C++ ```#include <bits/stdc++.h> using namespace std; class Node { public: int data; Node(int d) { data = d; } Node* ptr; }; // Function to check if the linked list // is palindrome or not bool isPalin(Node* head) { // Temp pointer Node* slow = head; // Declare a stack stack<int> s; // Push all elements of the list // to the stack while (slow != NULL) { s.push(slow->data); // Move ahead slow = slow->ptr; } // Iterate in the list again and // check by popping from the stack while (head != NULL) { // Get the top most element int i = s.top(); // Pop the element s.pop(); // Check if data is not // same as popped element if (head->data != i) { return false; } // Move ahead head = head->ptr; } return true; } // Driver Code int main() { // Addition of linked list Node one = Node(1); Node two = Node(2); Node three = Node(3); Node four = Node(2); Node five = Node(1); // Initialize the next pointer // of every current pointer five.ptr = NULL; one.ptr = &two; two.ptr = &three; three.ptr = &four; four.ptr = &five; Node* temp = &one; // Call function to check palindrome or not int result = isPalin(&one); if (result == 1) cout << "isPalindrome is true\n"; else cout << "isPalindrome is false\n"; return 0; } // This code has been contributed by Striver ``` Java ```/* Java program to check if linked list is palindrome * recursively */ import java.util.*; class linkedList { public static void main(String args[]) { Node one = new Node(1); Node two = new Node(2); Node three = new Node(3); Node four = new Node(4); Node five = new Node(3); Node six = new Node(2); Node seven = new Node(1); one.ptr = two; two.ptr = three; three.ptr = four; four.ptr = five; five.ptr = six; six.ptr = seven; boolean condition = isPalindrome(one); System.out.println("isPalidrome :" + condition); } static boolean isPalindrome(Node head) { Node slow = head; boolean ispalin = true; Stack<Integer> stack = new Stack<Integer>(); while (slow != null) { stack.push(slow.data); slow = slow.ptr; } while (head != null) { int i = stack.pop(); if (head.data == i) { ispalin = true; } else { ispalin = false; break; } head = head.ptr; } return ispalin; } } class Node { int data; Node ptr; Node(int d) { ptr = null; data = d; } } ``` C# ```// C# program to check if linked list // is palindrome recursively using System; using System.Collections.Generic; class linkedList { // Driver code public static void Main(String[] args) { Node one = new Node(1); Node two = new Node(2); Node three = new Node(3); Node four = new Node(4); Node five = new Node(3); Node six = new Node(2); Node seven = new Node(1); one.ptr = two; two.ptr = three; three.ptr = four; four.ptr = five; five.ptr = six; six.ptr = seven; bool condition = isPalindrome(one); Console.WriteLine("isPalidrome :" + condition); } static bool isPalindrome(Node head) { Node slow = head; bool ispalin = true; Stack<int> stack = new Stack<int>(); while (slow != null) { stack.Push(slow.data); slow = slow.ptr; } while (head != null) { int i = stack.Pop(); if (head.data == i) { ispalin = true; } else { ispalin = false; break; } head = head.ptr; } return ispalin; } } class Node { public int data; public Node ptr; public Node(int d) { ptr = null; data = d; } } // This code is contributed by amal kumar choubey ``` Javascript ```<script> /* JavaScript program to check if linked list is palindrome recursively */ class Node { constructor(val) { this.data = val; this.ptr = null; } } var one = new Node(1); var two = new Node(2); var three = new Node(3); var four = new Node(4); var five = new Node(3); var six = new Node(2); var seven = new Node(1); one.ptr = two; two.ptr = three; three.ptr = four; four.ptr = five; five.ptr = six; six.ptr = seven; var condition = isPalindrome(one); document.write("isPalidrome: " + condition); function isPalindrome(head) { var slow = head; var ispalin = true; var stack = []; while (slow != null) { stack.push(slow.data); slow = slow.ptr; } while (head != null) { var i = stack.pop(); if (head.data == i) { ispalin = true; } else { ispalin = false; break; } head = head.ptr; } return ispalin; } // This code is contributed by todaysgaurav </script> ``` Python3 ```# Python3 program to check if linked # list is palindrome using stack class Node: def __init__(self, data): self.data = data self.ptr = None # Function to check if the linked list # is palindrome or not def ispalindrome(head): # Temp pointer slow = head # Declare a stack stack = [] ispalin = True # Push all elements of the list # to the stack while slow != None: stack.append(slow.data) # Move ahead slow = slow.ptr # Iterate in the list again and # check by popping from the stack while head != None: # Get the top most element i = stack.pop() # Check if data is not # same as popped element if head.data == i: ispalin = True else: ispalin = False break # Move ahead head = head.ptr return ispalin # Driver Code # Addition of linked list one = Node(1) two = Node(2) three = Node(3) four = Node(4) five = Node(3) six = Node(2) seven = Node(1) # Initialize the next pointer # of every current pointer one.ptr = two two.ptr = three three.ptr = four four.ptr = five five.ptr = six six.ptr = seven seven.ptr = None # Call function to check palindrome or not result = ispalindrome(one) print("isPalindrome:", result) # This code is contributed by Nishtha Goel ```

Output
```isPalindrome is true
```

Time complexity: O(N), Iterating over the linked list of size N.
Auxiliary Space: O(N), Using an auxiliary stack

## Check if a Singly Linked List is Palindrome by Reversing the Linked List:

The idea is to first reverse the second half part of the linked list and then check whether the list is palindrome or not.

Follow the steps below to solve the problem:

• Get the middle of the linked list.Â
• Reverse the second half of the linked list.Â
• Check if the first half and second half are identical.Â
• Construct the original linked list by reversing the second half again and attaching it back to the first half

Below is the implementation of the above approach:

C++ ```// C++ program to check if a linked list is palindrome #include <bits/stdc++.h> using namespace std; // Link list node struct Node { char data; struct Node* next; Node(int x) { data = x; next = NULL; } }; void reverse(struct Node**); bool compareLists(struct Node*, struct Node*); // Function to check if given linked list is // palindrome or not bool isPalindrome(struct Node* head) { struct Node *slow_ptr = head, *fast_ptr = head; struct Node *second_half, *prev_of_slow_ptr = head; // To handle odd size list struct Node* midnode = NULL; // initialize result bool res = true; if (head != NULL && head->next != NULL) { // Get the middle of the list. Move slow_ptr by 1 // and fast_ptr by 2, slow_ptr will have the middle // node while (fast_ptr != NULL && fast_ptr->next != NULL) { fast_ptr = fast_ptr->next->next; // We need previous of the slow_ptr for // linked lists with odd elements prev_of_slow_ptr = slow_ptr; slow_ptr = slow_ptr->next; } // fast_ptr would become NULL when there // are even elements in list. And not NULL // for odd elements. We need to skip the // middle node for odd case and store it // somewhere so that we can restore the // original list if (fast_ptr != NULL) { midnode = slow_ptr; slow_ptr = slow_ptr->next; } // Now reverse the second half and // compare it with first half second_half = slow_ptr; // NULL terminate first half prev_of_slow_ptr->next = NULL; // Reverse the second half reverse(&second_half); // compare res = compareLists(head, second_half); // Construct the original list back reverse( &second_half); // Reverse the second half again // If there was a mid node (odd size case) // which was not part of either first half // or second half. if (midnode != NULL) { prev_of_slow_ptr->next = midnode; midnode->next = second_half; } else prev_of_slow_ptr->next = second_half; } return res; } // Function to reverse the linked list // Note that this function may change // the head void reverse(struct Node** head_ref) { struct Node* prev = NULL; struct Node* current = *head_ref; struct Node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head_ref = prev; } // Function to check if two input // lists have same data bool compareLists(struct Node* head1, struct Node* head2) { struct Node* temp1 = head1; struct Node* temp2 = head2; while (temp1 && temp2) { if (temp1->data == temp2->data) { temp1 = temp1->next; temp2 = temp2->next; } else return 0; } // Both are empty return 1 if (temp1 == NULL && temp2 == NULL) return 1; // Will reach here when one is NULL // and other is not return 0; } // Push a node to linked list. Note // that this function changes the head void push(struct Node** head_ref, char new_data) { // Allocate node struct Node* new_node = new Node(new_data); // Link the old list of the new node new_node->next = (*head_ref); // Move the head to point to the new node (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct Node* ptr) { while (ptr != NULL) { cout << ptr->data << "->"; ptr = ptr->next; } cout << "NULL" << "\n"; } // Driver code int main() { // Start with the empty list struct Node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); } isPalindrome(head) ? cout << "Is Palindrome" << "\n\n" : cout << "Not Palindrome" << "\n\n"; return 0; } // This code is contributed by Shivani ``` C ```/* C Program to check if a linked list is palindrome */ #include <stdbool.h> #include <stdio.h> #include <stdlib.h> /* Link list node */ struct Node { char data; struct Node* next; }; void reverse(struct Node**); bool compareLists(struct Node*, struct Node*); /* Function to check if given linked list is palindrome or not */ bool isPalindrome(struct Node* head) { struct Node *slow_ptr = head, *fast_ptr = head; struct Node *second_half, *prev_of_slow_ptr = head; struct Node* midnode = NULL; // To handle odd size list bool res = true; // initialize result if (head != NULL && head->next != NULL) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr will have the middle node */ while (fast_ptr != NULL && fast_ptr->next != NULL) { fast_ptr = fast_ptr->next->next; /*We need previous of the slow_ptr for linked lists with odd elements */ prev_of_slow_ptr = slow_ptr; slow_ptr = slow_ptr->next; } /* fast_ptr would become NULL when there are even elements in list. And not NULL for odd elements. We need to skip the middle node for odd case and store it somewhere so that we can restore the original list*/ if (fast_ptr != NULL) { midnode = slow_ptr; slow_ptr = slow_ptr->next; } // Now reverse the second half and compare it with // first half second_half = slow_ptr; prev_of_slow_ptr->next = NULL; // NULL terminate first half reverse(&second_half); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse( &second_half); // Reverse the second half again // If there was a mid node (odd size case) which // was not part of either first half or second half. if (midnode != NULL) { prev_of_slow_ptr->next = midnode; midnode->next = second_half; } else prev_of_slow_ptr->next = second_half; } return res; } /* Function to reverse the linked list Note that this function may change the head */ void reverse(struct Node** head_ref) { struct Node* prev = NULL; struct Node* current = *head_ref; struct Node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head_ref = prev; } /* Function to check if two input lists have same data*/ bool compareLists(struct Node* head1, struct Node* head2) { struct Node* temp1 = head1; struct Node* temp2 = head2; while (temp1 && temp2) { if (temp1->data == temp2->data) { temp1 = temp1->next; temp2 = temp2->next; } else return 0; } /* Both are empty return 1*/ if (temp1 == NULL && temp2 == NULL) return 1; /* Will reach here when one is NULL and other is not */ return 0; } /* Push a node to linked list. Note that this function changes the head */ void push(struct Node** head_ref, char new_data) { /* allocate node */ struct Node* new_node = (struct Node*)malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct Node* ptr) { while (ptr != NULL) { printf("%c->", ptr->data); ptr = ptr->next; } printf("NULL\n"); } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct Node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); } isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n"); return 0; } ``` Java ```/* Java program to check if linked list is palindrome */ class LinkedList { Node head; // head of list Node slow_ptr, fast_ptr, second_half; /* Linked list Node*/ class Node { char data; Node next; Node(char d) { data = d; next = null; } } /* Function to check if given linked list is palindrome or not */ boolean isPalindrome(Node head) { slow_ptr = head; fast_ptr = head; Node prev_of_slow_ptr = head; Node midnode = null; // To handle odd size list boolean res = true; // initialize result if (head != null && head.next != null) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr will have the middle node */ while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; /*We need previous of the slow_ptr for linked lists with odd elements */ prev_of_slow_ptr = slow_ptr; slow_ptr = slow_ptr.next; } /* fast_ptr would become NULL when there are even elements in the list and not NULL for odd elements. We need to skip the middle node for odd case and store it somewhere so that we can restore the original list */ if (fast_ptr != null) { midnode = slow_ptr; slow_ptr = slow_ptr.next; } // Now reverse the second half and compare it // with first half second_half = slow_ptr; prev_of_slow_ptr.next = null; // NULL terminate first half reverse(); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(); // Reverse the second half again if (midnode != null) { // If there was a mid node (odd size case) // which was not part of either first half // or second half. prev_of_slow_ptr.next = midnode; midnode.next = second_half; } else prev_of_slow_ptr.next = second_half; } return res; } /* Function to reverse the linked list Note that this function may change the head */ void reverse() { Node prev = null; Node current = second_half; Node next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } second_half = prev; } /* Function to check if two input lists have same data*/ boolean compareLists(Node head1, Node head2) { Node temp1 = head1; Node temp2 = head2; while (temp1 != null && temp2 != null) { if (temp1.data == temp2.data) { temp1 = temp1.next; temp2 = temp2.next; } else return false; } /* Both are empty return 1*/ if (temp1 == null && temp2 == null) return true; /* Will reach here when one is NULL and other is not */ return false; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list of the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { System.out.print(ptr.data + "->"); ptr = ptr.next; } System.out.println("NULL"); } /* Driver program to test the above functions */ public static void main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char str[] = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; String string = new String(str); for (int i = 0; i < 7; i++) { llist.push(str[i]); } if (llist.isPalindrome(llist.head) != false) { System.out.println("Is Palindrome"); System.out.println(""); } else { System.out.println("Not Palindrome"); System.out.println(""); } } } ``` C# ```/* C# program to check if linked list is palindrome */ using System; class LinkedList { Node head; // head of list Node slow_ptr, fast_ptr, second_half; /* Linked list Node*/ public class Node { public char data; public Node next; public Node(char d) { data = d; next = null; } } /* Function to check if given linked list is palindrome or not */ Boolean isPalindrome(Node head) { slow_ptr = head; fast_ptr = head; Node prev_of_slow_ptr = head; Node midnode = null; // To handle odd size list Boolean res = true; // initialize result if (head != null && head.next != null) { /* Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr will have the middle node */ while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; /*We need previous of the slow_ptr for linked lists with odd elements */ prev_of_slow_ptr = slow_ptr; slow_ptr = slow_ptr.next; } /* fast_ptr would become NULL when there are even elements in the list and not NULL for odd elements. We need to skip the middle node for odd case and store it somewhere so that we can restore the original list */ if (fast_ptr != null) { midnode = slow_ptr; slow_ptr = slow_ptr.next; } // Now reverse the second half and compare it // with first half second_half = slow_ptr; prev_of_slow_ptr.next = null; // NULL terminate first half reverse(); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(); // Reverse the second half again if (midnode != null) { // If there was a mid node (odd size case) // which was not part of either first half // or second half. prev_of_slow_ptr.next = midnode; midnode.next = second_half; } else prev_of_slow_ptr.next = second_half; } return res; } /* Function to reverse the linked list Note that this function may change the head */ void reverse() { Node prev = null; Node current = second_half; Node next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } second_half = prev; } /* Function to check if two input lists have same data*/ Boolean compareLists(Node head1, Node head2) { Node temp1 = head1; Node temp2 = head2; while (temp1 != null && temp2 != null) { if (temp1.data == temp2.data) { temp1 = temp1.next; temp2 = temp2.next; } else return false; } /* Both are empty return 1*/ if (temp1 == null && temp2 == null) return true; /* Will reach here when one is NULL and other is not */ return false; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list of the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { Console.Write(ptr.data + "->"); ptr = ptr.next; } Console.WriteLine("NULL"); } /* Driver program to test the above functions */ public static void Main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; for (int i = 0; i < 7; i++) { llist.push(str[i]); } if (llist.isPalindrome(llist.head) != false) { Console.WriteLine("Is Palindrome"); Console.WriteLine(""); } else { Console.WriteLine("Not Palindrome"); Console.WriteLine(""); } } } // This code is contributed by Arnab Kundu ``` Javascript ```<script> /* javascript program to check if linked list is palindrome */ var head; // head of list var slow_ptr, fast_ptr, second_half; /* Linked list Node */ class Node { constructor(val) { this.data = val; this.next = null; } } /* * Function to check if given linked list is palindrome or not */ function isPalindrome(head) { slow_ptr = head; fast_ptr = head; var prev_of_slow_ptr = head; var midnode = null; // To handle odd size list var res = true; // initialize result if (head != null && head.next != null) { /* * Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr * will have the middle node */ while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; /* * We need previous of the slow_ptr for linked lists with odd elements */ prev_of_slow_ptr = slow_ptr; slow_ptr = slow_ptr.next; } /* * fast_ptr would become NULL when there are even elements in the list and not * NULL for odd elements. We need to skip the middle node for odd case and store * it somewhere so that we can restore the original list */ if (fast_ptr != null) { midnode = slow_ptr; slow_ptr = slow_ptr.next; } // Now reverse the second half and compare it with first half second_half = slow_ptr; prev_of_slow_ptr.next = null; // NULL terminate first half reverse(); // Reverse the second half res = compareLists(head, second_half); // compare /* Construct the original list back */ reverse(); // Reverse the second half again if (midnode != null) { // If there was a mid node (odd size case) which // was not part of either first half or second half. prev_of_slow_ptr.next = midnode; midnode.next = second_half; } else prev_of_slow_ptr.next = second_half; } return res; } /* * Function to reverse the linked list Note that this function may change the * head */ function reverse() { var prev = null; var current = second_half; var next; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } second_half = prev; } /* Function to check if two input lists have same data */ function compareLists(head1, head2) { var temp1 = head1; var temp2 = head2; while (temp1 != null && temp2 != null) { if (temp1.data == temp2.data) { temp1 = temp1.next; temp2 = temp2.next; } else return false; } /* Both are empty return 1 */ if (temp1 == null && temp2 == null) return true; /* * Will reach here when one is NULL and other is not */ return false; } /* * Push a node to linked list. Note that this function changes the head */ function push( new_data) { /* * Allocate the Node & Put in the data */ var new_node = new Node(new_data); /* link the old list of the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list function printList(ptr) { while (ptr != null) { document.write(ptr.data + "->"); ptr = ptr.next; } document.write("NULL<br/>"); } /* Driver program to test the above functions */ /* Start with the empty list */ var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]; var string = str.toString(); for (i = 0; i < 7; i++) { push(str[i]); printList(head); if (isPalindrome(head) != false) { document.write("Is Palindrome"); document.write("<br/>"); } else { document.write("Not Palindrome"); document.write("<br/>"); } } // This code contributed by gauravrajput1 </script> ``` Python3 ```# Python3 program to check if # linked list is palindrome # Node class class Node: # Constructor to initialize # the node object def __init__(self, data): self.data = data self.next = None class LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to check if given # linked list is palindrome or not def isPalindrome(self, head): slow_ptr = head fast_ptr = head prev_of_slow_ptr = head # To handle odd size list midnode = None # Initialize result res = True if (head != None and head.next != None): # Get the middle of the list. # Move slow_ptr by 1 and # fast_ptr by 2, slow_ptr # will have the middle node while (fast_ptr != None and fast_ptr.next != None): # We need previous of the slow_ptr # for linked lists with odd # elements fast_ptr = fast_ptr.next.next prev_of_slow_ptr = slow_ptr slow_ptr = slow_ptr.next # fast_ptr would become NULL when # there are even elements in the # list and not NULL for odd elements. # We need to skip the middle node for # odd case and store it somewhere so # that we can restore the original list if (fast_ptr != None): midnode = slow_ptr slow_ptr = slow_ptr.next # Now reverse the second half # and compare it with first half second_half = slow_ptr # NULL terminate first half prev_of_slow_ptr.next = None # Reverse the second half second_half = self.reverse(second_half) # Compare res = self.compareLists(head, second_half) # Construct the original list back # Reverse the second half again second_half = self.reverse(second_half) if (midnode != None): # If there was a mid node (odd size # case) which was not part of either # first half or second half. prev_of_slow_ptr.next = midnode midnode.next = second_half else: prev_of_slow_ptr.next = second_half return res # Function to reverse the linked list # Note that this function may change # the head def reverse(self, second_half): prev = None current = second_half next = None while current != None: next = current.next current.next = prev prev = current current = next second_half = prev return second_half # Function to check if two input # lists have same data def compareLists(self, head1, head2): temp1 = head1 temp2 = head2 while (temp1 and temp2): if (temp1.data == temp2.data): temp1 = temp1.next temp2 = temp2.next else: return 0 # Both are empty return 1 if (temp1 == None and temp2 == None): return 1 # Will reach here when one is NULL # and other is not return 0 # Function to insert a new node # at the beginning def push(self, new_data): # Allocate the Node & # Put in the data new_node = Node(new_data) # Link the old list of the new one new_node.next = self.head # Move the head to point to new Node self.head = new_node # A utility function to print # a given linked list def printList(self): temp = self.head while(temp): print(temp.data, end="->") temp = temp.next print("NULL") # Driver code if __name__ == '__main__': l = LinkedList() s = ['a', 'b', 'a', 'c', 'a', 'b', 'a'] for i in range(7): l.push(s[i]) if (l.isPalindrome(l.head) != False): print("Is Palindrome\n") else: print("Not Palindrome\n") # This code is contributed by MuskanKalra1 ```

Output
```Is Palindrome

```

Time Complexity: O(N),Â Â
Auxiliary Space: O(1)

## Check if a Singly Linked List is Palindrome using Recursion:Â

The idea is to use the function call stack as a container. Recursively traverse till the end of the list. When returning from the last NULL, will be at the last node. The last node is to be compared with the first node of the list.

Follow the steps below to solve the problem:

• First make a recursive call to the next node of linked till it reach the last node.
• After returning from last node start checking from start of the linked list then move to the next node.
• Repeat these steps till reach the last node.

Below is the implementation of above approach:

C++ ```// Recursive program to check if a given linked list is // palindrome #include <bits/stdc++.h> using namespace std; /* Link list node */ struct Node { char data; struct Node* next; Node(int x) { data = x; next = NULL; } }; // Initial parameters to this function are &head and head bool isPalindromeUtil(struct Node** left, struct Node* right) { /* stop recursion when right becomes NULL */ if (right == NULL) return true; /* If sub-list is not palindrome then no need to check for current left and right, return false */ bool isp = isPalindromeUtil(left, right->next); if (isp == false) return false; /* Check values at current left and right */ bool isp1 = (right->data == (*left)->data); /* Move left to next node */ *left = (*left)->next; return isp1; } // A wrapper over isPalindromeUtil() bool isPalindrome(struct Node* head) { return isPalindromeUtil(&head, head); } /* Push a node to linked list. Note that this function changes the head */ void push(struct Node** head_ref, char new_data) { /* allocate node */ struct Node* new_node = new Node(new_data); /* put in the data */ new_node->data = new_data; /* link the old list of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct Node* ptr) { while (ptr != NULL) { cout << ptr->data << "->"; ptr = ptr->next; } cout << "NULL\n"; } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct Node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); } isPalindrome(head) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n"; return 0; } // this code is contributed by shivanisinghss2110 ``` C ```// Recursive program to check if a given linked list is // palindrome #include <stdbool.h> #include <stdio.h> #include <stdlib.h> /* Link list node */ struct node { char data; struct node* next; }; // Initial parameters to this function are &head and head bool isPalindromeUtil(struct node** left, struct node* right) { /* stop recursion when right becomes NULL */ if (right == NULL) return true; /* If sub-list is not palindrome then no need to check for current left and right, return false */ bool isp = isPalindromeUtil(left, right->next); if (isp == false) return false; /* Check values at current left and right */ bool isp1 = (right->data == (*left)->data); /* Move left to next node */ *left = (*left)->next; return isp1; } // A wrapper over isPalindromeUtil() bool isPalindrome(struct node* head) { return isPalindromeUtil(&head, head); } /* Push a node to linked list. Note that this function changes the head */ void push(struct node** head_ref, char new_data) { /* allocate node */ struct node* new_node = (struct node*)malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } // A utility function to print a given linked list void printList(struct node* ptr) { while (ptr != NULL) { printf("%c->", ptr->data); ptr = ptr->next; } printf("NULL\n"); } /* Driver program to test above function*/ int main() { /* Start with the empty list */ struct node* head = NULL; char str[] = "abacaba"; int i; for (i = 0; str[i] != '\0'; i++) { push(&head, str[i]); } isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n"); return 0; } ``` Java ```// Java program for the above approach public class LinkedList { // Head of the list Node head; Node left; public class Node { public char data; public Node next; // Linked list node public Node(char d) { data = d; next = null; } } // Initial parameters to this function are // &head and head boolean isPalindromeUtil(Node right) { left = head; // Stop recursion when right becomes null if (right == null) return true; // If sub-list is not palindrome then no need to // check for the current left and right, return // false boolean isp = isPalindromeUtil(right.next); if (isp == false) return false; // Check values at current left and right boolean isp1 = (right.data == left.data); left = left.next; // Move left to next node; return isp1; } // A wrapper over isPalindrome(Node head) boolean isPalindrome(Node head) { boolean result = isPalindromeUtil(head); return result; } // Push a node to linked list. Note that // this function changes the head public void push(char new_data) { // Allocate the node and put in the data Node new_node = new Node(new_data); // Link the old list of the new one new_node.next = head; // Move the head to point to new node head = new_node; } // A utility function to print a // given linked list void printList(Node ptr) { while (ptr != null) { System.out.print(ptr.data + "->"); ptr = ptr.next; } System.out.println("Null"); } // Driver Code public static void main(String[] args) { LinkedList llist = new LinkedList(); char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; for (int i = 0; i < 7; i++) { llist.push(str[i]); } if (llist.isPalindrome(llist.head)) { System.out.println("Is Palindrome"); System.out.println(""); } else { System.out.println("Not Palindrome"); System.out.println(""); } } } // This code is contributed by abhinavjain194 ``` C# ```/* C# program to check if linked list is palindrome recursively */ using System; public class LinkedList { Node head; // head of list Node left; /* Linked list Node*/ public class Node { public char data; public Node next; public Node(char d) { data = d; next = null; } } // Initial parameters to this function are &head and // head Boolean isPalindromeUtil(Node right) { left = head; /* stop recursion when right becomes NULL */ if (right == null) return true; /* If sub-list is not palindrome then no need to check for current left and right, return false */ Boolean isp = isPalindromeUtil(right.next); if (isp == false) return false; /* Check values at current left and right */ Boolean isp1 = (right.data == (left).data); /* Move left to next node */ left = left.next; return isp1; } // A wrapper over isPalindromeUtil() Boolean isPalindrome(Node head) { Boolean result = isPalindromeUtil(head); return result; } /* Push a node to linked list. Note that this function changes the head */ public void push(char new_data) { /* Allocate the Node & Put in the data */ Node new_node = new Node(new_data); /* link the old list of the new one */ new_node.next = head; /* Move the head to point to new Node */ head = new_node; } // A utility function to print a given linked list void printList(Node ptr) { while (ptr != null) { Console.Write(ptr.data + "->"); ptr = ptr.next; } Console.WriteLine("NULL"); } /* Driver code */ public static void Main(String[] args) { /* Start with the empty list */ LinkedList llist = new LinkedList(); char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' }; // String string = new String(str); for (int i = 0; i < 7; i++) { llist.push(str[i]); } if (llist.isPalindrome(llist.head) != false) { Console.WriteLine("Is Palindrome"); Console.WriteLine(""); } else { Console.WriteLine("Not Palindrome"); Console.WriteLine(""); } } } // This code is contributed by Rajput-Ji ``` Javascript ```<script> // javascript program for the above approach // Head of the list var head; var left; class Node { constructor(val) { this.data = val; this.next = null; } } // Initial parameters to this function are // &head and head function isPalindromeUtil( right) { left = head; // Stop recursion when right becomes null if (right == null) return true; // If sub-list is not palindrome then no need to // check for the current left and right, return // false var isp = isPalindromeUtil(right.next); if (isp == false) return false; // Check values at current left and right var isp1 = (right.data == left.data); left = left.next; // Move left to next node; return isp1; } // A wrapper over isPalindrome(Node head) function isPalindrome( head) { var result = isPalindromeUtil(head); return result; } // Push a node to linked list. Note that // this function changes the head function push( new_data) { // Allocate the node and put in the data var new_node = new Node(new_data); // Link the old list of the new one new_node.next = head; // Move the head to point to new node head = new_node; } // A utility function to print a // given linked list function printList( ptr) { while (ptr != null) { document.write(ptr.data + "->"); ptr = ptr.next; } document.write("Null "); document.write("<br>"); } // Driver Code var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]; for (var i = 0; i < 7; i++) { push(str[i]); printList(head); if (isPalindrome(head)) { document.write("Is Palindrome"); document.write("<br/>"); document.write("<br>"); } else { document.write("Not Palindrome"); document.write("<br/>"); document.write("<br/>"); } } // This code contributed by aashish1995 </script> ``` Python3 ```# Python program for the above approach # Head of the list head = None left = None class Node: def __init__(self, val): self.data = val self.next = None # Initial parameters to this function are # &head and head def isPalindromeUtil(right): global head, left left = head # Stop recursion when right becomes null if (right == None): return True # If sub-list is not palindrome then no need to # check for the current left and right, return # false isp = isPalindromeUtil(right.next) if (isp == False): return False # Check values at current left and right isp1 = (right.data == left.data) left = left.next # Move left to next node; return isp1 # A wrapper over isPalindrome(Node head) def isPalindrome(head): result = isPalindromeUtil(head) return result # Push a node to linked list. Note that # this function changes the head def push(new_data): global head # Allocate the node and put in the data new_node = Node(new_data) # Link the old list of the new one new_node.next = head # Move the head to point to new node head = new_node # A utility function to print a # given linked list def printList(ptr): while (ptr != None): print(ptr.data, end="->") ptr = ptr.next print("Null ") # Driver Code str = ['a', 'b', 'a', 'c', 'a', 'b', 'a'] for i in range(0, 7): push(str[i]) if (isPalindrome(head) and i != 0): print("Is Palindrome\n") else: print("Not Palindrome\n") # This code is contributed by saurabh_jaiswal. ```

Output
```Is Palindrome

```

Time Complexity: O(N), Traversing over the linked list of size N.
Auxiliary Space: O(N) if Function Call Stack size is considered, otherwise O(1).

Thanks to Sharad Chandra for suggesting this approach.Â  Â