# C++ Program To Check If Two Linked Lists Are Identical

• Last Updated : 03 Aug, 2022

Two Linked Lists are identical when they have the same data and the arrangement of data is also the same. For example, Linked lists a (1->2->3) and b(1->2->3) are identical. . Write a function to check if the given two linked lists are identical.

Method 1 (Iterative):
To identify if two lists are identical, we need to traverse both lists simultaneously, and while traversing we need to compare data.

## C++

 `// An iterative C++ program to check if ` `// two linked lists are identical or not` `#include` `using` `namespace` `std;`   `// Structure for a linked list node ` `struct` `Node` `{` `    ``int` `data;` `    ``struct` `Node *next;` `};`   `/* Returns true if linked lists a and b ` `   ``are identical, otherwise false */` `bool` `areIdentical(``struct` `Node *a, ` `                  ``struct` `Node *b)` `{` `    ``while` `(a != NULL && b != NULL)` `    ``{` `        ``if` `(a->data != b->data)` `            ``return` `false``;`   `        ``/* If we reach here, then a and b are ` `           ``not NULL and their data is same, so ` `           ``move to next nodes in both lists */` `        ``a = a->next;` `        ``b = b->next;` `    ``}`   `    ``// If linked lists are identical, then ` `    ``// 'a' and 'b' must be NULL at this point.` `    ``return` `(a == NULL && b == NULL);` `}`   `/* UTILITY FUNCTIONS TO TEST fun1() ` `   ``and fun2() */` `/* Given a reference (pointer to pointer) ` `   ``to the head of a list and an int, push ` `   ``a new node on the front of the list. */` `void` `push(``struct` `Node** head_ref, ` `          ``int` `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 point to the ` `    ``// new node ` `    ``(*head_ref) = new_node;` `}`   `// Driver Code` `int` `main()` `{` `    ``/* The constructed linked lists are :` `       ``a: 3->2->1` `       ``b: 3->2->1 */` `    ``struct` `Node *a = NULL;` `    ``struct` `Node *b = NULL;` `    ``push(&a, 1);` `    ``push(&a, 2);` `    ``push(&a, 3);` `    ``push(&b, 1);` `    ``push(&b, 2);` `    ``push(&b, 3);`   `    ``if``(areIdentical(a, b))` `        ``cout << ``"Identical"``;` `    ``else` `        ``cout << ``"Not identical"``;`   `    ``return` `0;` `}` `// This code is contributed by Akanksha Rai`

Output:

`Identical`

Method 2 (Recursive):
Recursive solution code is much cleaner than iterative code. You probably wouldn’t want to use the recursive version for production code, however, because it will use stack space which is proportional to the length of the lists.

## C++

 `// A recursive C++ function to check if two ` `// linked lists are identical or not ` `bool` `areIdentical(Node *a, Node *b) ` `{ ` `    ``// If both lists are empty ` `    ``if` `(a == NULL && b == NULL) ` `    ``return` `true``; `   `    ``// If both lists are not empty, then ` `    ``// data of current nodes must match, ` `    ``// and same should be recursively true ` `    ``// for rest of the nodes. ` `    ``if` `(a != NULL && b != NULL) ` `    ``return` `(a->data == b->data) && ` `            ``areIdentical(a->next, b->next); `   `    ``// If we reach here, then one of the lists ` `    ``// is empty and other is not ` `    ``return` `false``; ` `} ` `//This is code is contributed by rathbhupendra`

Time Complexity: O(n) for both iterative and recursive versions. n is the length of the smaller list among a and b.

Auxiliary Space: O(n) for call stack because using recursion