Linked List Interview Code methods, runtime, and edge cases

Refactored code here

I want to write a very simple Linked List with only 3 methods Append, Remove and Print. This is not for production and is to be treated as code that could be used in an interview or a quick and dirty prototype. I'm really curious about the approach I've taken here to ensure no duplicates appear in my Linked List. I feel the data structure will be much easier to work with if remove any complexity around duplicate data and want to use this Linked List to implement a Stack, or Queue, or Binary Search Tree etc. I had int data before as the member data field for my Linked List and don't want to make this overly complex by introducing a concept of ids.

First I'd like to know if my member functions for the Linked List have any edge cases I am not catching and any improvements I can make to run time efficiency . Anyway I can simplify this code further with c++11 features, shorter variable names or any other suggestions would be appreciated too.

#include <iostream>

using namespace std;

struct Node
{
    int id;
    Node* next;

    Node(int id) : id(id), next(nullptr) { }

    void append(int newId) {
        Node* current = this;
        while (current->next != nullptr) {
            if (current->id == newId) {
                return;
            }
            else {
                current = current->next;
            }
        }
        current->next = new Node(newId);
    }

    void remove(int targetId) {
        Node* current = this;
        Node* previous;
        while (current->id != targetId) {
            if (current->next != nullptr) {
                previous = current;
                current = current->next;
            }
            else {
                cout << "node not found :(\n";
                return;
            }
        }
        if (current->next == nullptr) {
            delete current;
            previous->next = nullptr;
        }
        else {
            Node* danglingPtr = current->next;
            current->id = current->next->id;
            current->next = current->next->next;
            delete danglingPtr;
        }
    }

     void print() {
        if (this->next == nullptr) {
            cout << this->id << endl;
        }
        else {
            cout << this->id << " ";
            this->next->print();
        }
    }
};

int main()
{
    Node list(1);

    list.append(2);
    list.append(3);
    list.append(4);
    list.print();

    list.remove(3);
    list.print();

    list.remove(4);
    list.print();

    list.remove(1337);
}