指向根的二叉树指针需要被引用和取消引用。 为什么？

Question

My question is why do I need to dereference and reference a pointer for the following code to work? Doesn't ref/deref cancel each other? I would really appreciate if anyone could explain it like I'm five :)

Code:

template <typename T> 
class binNode {
private:
    T key;
public:
    binNode * left;
    binNode * right;
    binNode * parent;
    binNode() {
        this->left = NULL;
        this->right = NULL;
        this->parent = NULL;
    }
    // arg constructor:
    binNode (T key) {
        this->key = key;
        this->left = NULL;
        this->right = NULL;
        this->parent = NULL;
    }

    T getKey() {
        return this->key;
    }
    void setKey(T key) {
        this->key = key;
    }
};

template<typename T> class Tree {
private:
    binNode <T> *root;
public:
    Tree() {
        this->root = NULL;
    }
    Tree(binNode <T> * node) {
        node->parent = NULL;
        this->root = node;
    }
    /* THIS IS THE PART I  DON'T GET */
    void addNode(binNode<T> *&x, binNode<T> * node) { // what's up with the *&???
        if (x == NULL) {
            x = node;
            return;
        } else if (x->getKey() == node->getKey()) {
            node->left = x;
            node->parent = x->parent;
            x->parent = node;
            return;
        }

        if (node->getKey() < x->getKey()) {
            addNode(x->left, node);
        } else {
            addNode(x->right, node);
        }

    }

    void addNode(binNode<T> * node) {
        addNode(this->root, node);
    }

    binNode<T> * treeSearch(binNode<T> * x, T key) {
        if (x == NULL || key == x->getKey()) {
            return x;
        }
        if (key < x->getKey()) {
            return treeSearch(x->left, key);
        } else {
            return treeSearch(x->right, key);
        }
    }

    void printOrdered() {
        inorderTreeWalk(root);
        cout << endl;
    }

    void inorderTreeWalk(binNode<T> * node) {
        if (node != NULL) {
            inorderTreeWalk(node->left);
            cout << node->getKey() << '\t';
            inorderTreeWalk(node->right);
        }
    }

};

Here is the main function ( #inlude is not included)

int main() {
    Tree<int> T (new binNode<int>(10));
    // Tree<int> T = new binNode<int>(10);

    T.addNode(new binNode<int> (11));
    T.addNode(new binNode<int> (9));
    T.addNode(new binNode<int> (8));
    T.addNode(new binNode<int> (12));

    T.printOrdered();

}

Answer 1

That's not a reference / dereference of a pointer , it's a reference to a pointer. It is necessary because...

void addNode(binNode<T> *&x, binNode<T> * node) {
    if (x == NULL) {
        x = node; // ...here...
        return;
    } else // ...

...you are assigning to the parameter x .

If you hadn't passed the pointer x by reference , you would assign to the local copy of the parameter:

void addNode(binNode<T> * x, binNode<T> * node) {
    if (x == NULL) {
        x = node; // this acts on the local copy only, and thus does nothing.
        return;
    } else // ...

Answer 2

Via the pointer (without the reference), you get a local copy of the address. Which means you can manipulate the value behind the pointer (in this case *x) which would change. But if you change the address itself, the address would behave like a local copy and you lose the address-changes after leaving the method.