Strange way of printing Binary Search Tree C++ with pointers

Question

I was going through a lecture and it showed some code that prints out a binary search tree recursively like this

void printTree(node *t){
    if(t!=NULL){
        printTree(t->left);
        cout<<t->key<<endl;
        printTree(t->right);
     }
 }

I understand what it is doing but I don't understand the pointers. The function is passing a pointer to a node yet in the 'cout' line, it is trying to access the key value in the node struct without dereferencing it first. What I mean is, shouldn't it be something like

cout<<(*t)->key<<endl;

instead?

Answer 1

Actually, -> is a dereferencing operator. You can make a choice:

cout<<t->key<<endl; or cout<<(*t).key<<endl;

Answer 2

但是绝对不是“（* t）-> key”-这将是双重deref，如果不是崩溃，可能是编译错误。