Question

I have a Binary Search Tree that I’m making and I implemented the Insert node code as follows:

BSTNode * BST::Insert(const std::string & v){
    BSTNode * node = new BSTNode(v);
    if (root == NULL){
        root = node;
    } else if (root->value.compare(v) > 0) {
        Insert(root->left, node);
        //recursive insert the method
    } else if (root->value.compare(v) < 0) {
        Insert(root->right, node);
    } else {
        delete node;
        return NULL;
    }
    size++;
    return node;
}

Followed by the recursive insert method in my header file (it has private access):

void Insert(BSTNode* current, BSTNode* node){
    if (current == NULL){
        current = node;
    } else if (current->value == node->value){
        delete node;
    } else if (current->value < node->value) {
        Insert(current->left, node);
    } else if (current->value > node->value){//if (parent->value < node->value) {
        Insert(current->right, node);
    }
}

When the recursive function sets the pointer and then returns, the changes I made to the pointer DON’T stay.

Answer 1

The problem here is that current is a local variable with a copy of the value of the pointer. It’s not the same variable that you pass in. If you want to modify the pointer in place, your method should accept either a pointer to the pointer or a reference to the pointer. The easiest way to do this would be to just modify current to be a reference to a pointer. The resultant code would look like this:

void Insert(BSTNode* &current, BSTNode* node){
    if (current == NULL){
        current = node;
    } else if (current->value == node->value){
        delete node;
    } else if (current->value < node->value) {
        Insert(current->left, node);
    } else if (current->value > node->value){//if (parent->value < node->value) {
        Insert(current->right, node);
    }
}

In general, you should keep in mind that pointers are just values which tell you the memory location of something. So they’re really just numbers. And when they’re passed into a function, unless there’s a reference, they’re passed by the value. So your function just sees the number, not the variable which contained the number.