C++ Binary Tree Printing Nodes

Question

I'm learning about binary trees. I was looking at Stanford website: http://cslibrary.stanford.edu/110/BinaryTrees.html There was a practice problem of making a tree by calling newNode() three times, and using three pointer variables. The struct and newNode were given. I was trying to print out the nodes.

struct node { 
    int data; 
    struct node* left; 
    struct node* right; 
} ;

/* 
 Helper function that allocates a new node 
 with the given data and NULL left and right pointers. 
*/ 
struct node* newNode(int data) { 
  struct node* node = new(struct node); 
  node->data = data; 
  node->left = NULL; 
  node->right = NULL;

  return(node); 
}; 

// call newNode() three times 
struct node* build123a() { 
  struct node* root = newNode(2); 
  struct node* lChild = newNode(1); 
  struct node* rChild = newNode(3);
  root->left = lChild; 
  root->right= rChild;

  return(root); 
}

int main() {

    struct node* test = build123a();
    cout << "root: " << test->data << endl;
    cout << "left: " << test->left << endl;
    cout << "right: " << test->right << endl;

    return 0;
}

The issue is that this prints out only the integer in root. For the left and right nodes, it prints out address locations. My knowledge of pointers is still a little shaky. But it shouldn't matter that I've only returned root right? newNode is a pointer to a node right? Just looking for a simple fix to print out left and right nodes.

Answer 1

That's because 'left' & 'right' are pointers.

To print out the 'data' of the left or right, change the code as follows:

cout << "left: " << test->left->data << endl;

cout << "right: " << test->right->data << endl;

Note, however, if left or right are NULL (ie, zero) you'll likely get a memory access exception.

Answer 2

struct node { 
    int data; // the actual data contained inside this node
    struct node* left; // a node pointer that points to the left child
    struct node* right; // a node pointer that points to the right child
};

struct node* test; // is a node pointer
test->left; // is a node pointer that points to the left child of test
test->right; // is a node pointer that points to the right child of test

cout << test->data; // prints the integer contained within the test node
cout << test->left; // prints the address of the left child of test since it's a pointer
cout << test->right; // prints the address of the right child of test since it's a pointer

What you want to do is print the data contained within the left and right children.

cout << test->left->data;
cout << test->right->data;

Answer 3

test->left is (*test).left which is of type struct node* .

To print the data in left you need

cout << (test -> left -> data);

Answer 4

You can print "test->data" correctly because that's an int. The issue is that "test->left" and "test->right" are pointers, and pointers are basically numbers that refer to where another object is stored.

If you wanted to print the left node's data, you'd have to do this:

cout << "left: " << test->left->data << endl;

And then you'd have to do the same for the right node.