Function to check if a singly linked list is palindrome

Given a singly linked list of characters, write a function that returns true if the given list is a palindrome, else false.

Palindrome Linked List

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

METHOD 1 (Use a Stack)

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 stack and compare data of popped node with currently visited node.
If all nodes matched, then return true, else false.

Below image is a dry run of the above approach:

Below is the implementation of the above approach :

cpp

#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 true\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 linkeList { 
    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; 
    } 
} 

Output

 isPalindrome: true

The time complexity of the above method is O(n).

METHOD 2 (By reversing the list)
This method takes O(n) time and O(1) extra space.
1) Get the middle of the linked list.
2) Reverse the second half of the linked list.
3) Check if the first half and second half are identical.
4) Construct the original linked list by reversing the second half again and attaching it back to the first half

To divide the list in two halves, method 2 of this post is used.
When number of nodes are even, the first and second half contain exactly half nodes. The challenging thing in this method is to handle the case when number of nodes are odd. We don’t want the middle node as part of any of the lists as we are going to compare them for equality. For odd case, we use a separate variable ‘midnode’.

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_ptrr 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 reurn 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 off the new node */
    new_node->next = (*head_ref); 
  
    /* move the head to pochar 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"); 
} 
  
/* Drier 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]); 
        printList(head); 
        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_ptrr 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 reurn 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 off 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]); 
            llist.printList(llist.head); 
            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_ptrr 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 reurn 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 off 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]); 
            llist.printList(llist.head); 
            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 

Output:

a->NULL
Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome

Time Complexity: O(n)
Auxiliary Space: O(1)

METHOD 3 (Using Recursion)
Use two pointers left and right. Move right and left using recursion and check for following in each recursive call.
1) Sub-list is palindrome.
2) Value at current left and right are matching.

If both above conditions are true then return true.

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

In order to access first node of list, we need list head to be available in the last call of recursion. Hence we pass head also to the recursive function. If they both match we need to compare (2, n-2) nodes. Again when recursion falls back to (n-2)nd node, we need reference to 2nd node from head. We advance the head pointer in previous call, to refer to next node in the list.

However, the trick in identifying double pointer. Passing single pointer is as good as pass-by-value, and we will pass the same pointer again and again. We need to pass the address of head pointer for reflecting the changes in parent recursive calls.

Thanks to Sharad Chandra for suggesting this 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; 
}; 
  
// 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) 
{ 
    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 off the new node */
    new_node->next = (*head_ref); 
  
    /* move the head to pochar 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]); 
        printList(head); 
        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) 
{ 
    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 off the new node */
    new_node->next = (*head_ref); 
  
    /* move the head to pochar 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]); 
        printList(head); 
        isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n"); 
    } 
  
    return 0; 
} 

Java

/* Java program to check if linked list is palindrome recursively */
  
class LinkedList { 
    Node head; // head of list 
    Node left; 
  
    /* Linked list Node*/
    class Node { 
        char data; 
        Node next; 
  
        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 off 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]); 
            llist.printList(llist.head); 
            if (llist.isPalindrome(llist.head) != false) { 
                System.out.println("Is Palindrome"); 
                System.out.println(""); 
            } 
            else { 
                System.out.println("Not Palindrome"); 
                System.out.println(""); 
            } 
        } 
    } 
} 
  
// This code has been contributed by Mayank Jaiswal(mayank_24) 

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 off 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]);  
            llist.printList(llist.head);  
            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 

Output:

a->NULL
Not Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome

Time Complexity: O(n)
Auxiliary Space: O(n) if Function Call Stack size is considered, otherwise O(1).

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