C++ Binary Search Tree template return Node from function

Question

I've got the following BinaryTree class that includes a private class called Node (I've omitted everything except what's needed for this question):

template<typename T>
class BinaryTree{
private:
    template<typename NT>
    class Node{
    public:
        Node<NT>* left;
        Node<NT>* right;
        NT item;
    };
public:
    Node<T> Find(T itemToFind);
};

which is fairly standard. I'm trying to implement the Node FindMax() function that returns a Node object and I'm unable to figure out how to define it. For example, I assumed (incorrectly) that this would suffice:

template<typename T>
Node<T> BinaryTree<T>::Find(T itemToFind){ // -------> error line
    //...do something....
}

but I get an error saying: No template named Node . I've tried various other combinations to no avail. I'm not looking to return a bool if found, I want to return the Node itself.

Answer 1

You need to add scope operator :: since the Node class is part of the BinaryTree class:

template<typename T>
BinaryTree<T>::Node<T> BinaryTree<T>::Find(T itemToFind) {
    //...do something....
}

C++14 introduced auto return type so the following will work too:

template<typename T>
auto BinaryTree<T>::Find(T itemToFind) {
    //...do something....
}

Answer 2

At that point, the Node class is not in scope. You need to do something like this:

template<typename T>
BinaryTree<T>::Node<T> BinaryTree<T>::Find(T itemToFind) { 
    //...do something....
}

This was one of the motivations for C++11's trailing return types, which allow you to shorten the return type. IIRC, the trailing return type for member functions is evaluated within the scope of the class.

template<typename T>
BinaryTree<T>::Find(T itemToFind) -> Node<T> { 
    //...do something....
}

By the way, do you really want the Node class to have a template parameter that is independent of the template parameter for BinaryTree ? That is, you may want to eliminate the second template parameter NT .