binary trees

[StGuru]

I got one code with binary trees.
http://pastebin.com/f307cbe21

Is here:

void treeList(TreeNode *node) {
       // Print the items in the tree in postorder, one item
       // to a line.  Since the tree is a sort tree, the output
       // will be in increasing order.
   if ( node != NULL ) {
        treeList((*node).left);           // Print items in left subtree.
       cout << "  " << node->item << endl;  // Print item in the node.
       treeList(node->right);          // Print items in the right subtree.
   }
} // end treeList()

pointer pointing at pointer (*node).left or node->left ???
Thanks in advance.

[Lindley]

(*node).left or node->left

Those two expressions are equivalent.

[StGuru]

Yes I know. But is it pointer node pointing at pointer left of the struct TreeNode?

[Lindley]

If I'm parsing your question right, no, it doesn't modify what "node" points to.

[StGuru]

If left is not pointer, then what is it?

Thanks.

[StGuru]

For first time I see pointer pointing to pointer :-). I understand the binary trees theoretically, but I don't understand even single line of the code.

Thanks in advance.

[Lindley]

I'm not really sure what your question is anymore.

[StGuru]

Anyway, there are questions. I really don't understand the code. Can somebody explain?

[Lindley]

It's simply an inorder traversal of the entire tree....nothing particularly complicated there. You can read it as

IF (at a valid node)
    Output all values in my left child
    Output my value
    Output all values in my right child
END IF

Where the first and third outputs are recursive calls, which I know will output their subtrees' values because that's what I do to my own value in the second output.

[StGuru]

I don't understand the output. How does it come to node->item? There is treeList((*node).left); Which is pointer that points to another pointer "left". I actually don't understand how the pointers *left and *right store the items.

[Lindley]

http://en.wikipedia.org/wiki/Binary_tree
http://en.wikipedia.org/wiki/File:Binary_tree.svg

In that image, the root node passed to treeList() initially would be the one with node->item == 2. Then node->left->item == 7, node->right->item == 5, node->left->right->item == 6, etc.

[StGuru]

Thank you. Now I understand. When does it stop? Is it when the pointer is null or there isn't any more nodes in the tree?
Is it same like pointer pointing at pointer?

#include <iostream>
using namespace std;

int main()
{
   int a;
int * b;
int ** c;
a = 5;
b = &a;
c = &b;

cout<<a<<" "<<*b<<" "<<**c<<endl;
system("PAUSE");
}

[Lindley]

Typically such a tree would be built with the new operator, not with &. A NULL child pointer indicates no child node on that side.