solution

function reversePrint(llist) { // Write your code here if (llist === null) { return } else { // reverse the list let current = llist; let prev = null;

    while (current !== null) {
        let next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    };

    //print the reverse list
    let reversed = prev;
    for (; reversed !== null; reversed = reversed.next) {
        console.log(reversed.data);
    };

}

}

Here is my c++ solution, you can watch the vidéo explanation here : https://youtu.be/WO7Uz7sQML4

void reversePrint(SinglyLinkedListNode* llist) {
    if(llist == nullptr) return;
    reversePrint(llist->next);
    cout << llist->data << endl;
}

Here is my recursive call on next node for JAVA

public static void reversePrint(SinglyLinkedListNode llist) {
    if(llist == null) {
        return;
    }

    reversePrint(llist.next);

    System.out.println(llist.data);
}

Python recursion Solution def reversePrint(llist): if llist.next is None: print(llist.data) else: reversePrint(llist.next) print(llist.data)

Did it with simple vector in C++

void reversePrint(SinglyLinkedListNode* llist) {
    vector<int> arr;
    SinglyLinkedListNode* temp = llist;
    while(temp != NULL){
        arr.push_back(temp->data);
        temp = temp->next;
    }
    for(int i = arr.size() - 1; i >= 0; i--){
        cout << arr[i] << endl;
    }
}