Binary trees, returning the next node in a preorder traversal

Question

I have an assignment and I need some help with a method.

So I have a tree like this:

                A
              /   \
             B     C
           /   \ /   \
          D    E F     G
        /   \
       H      I
     /   \
   J       K

and my method is:

public BinaryTree preorderNext(BinaryTree t, BinaryTree v, BinaryTree prev) {       

    prev = t;

    if(t.getLeft() != null) {
        preorderNext(t.getLeft(), v, prev);
    }

    if(t.getRight() != null) {
        preorderNext(t.getRight(), v, prev);
    }


    if(prev == v) {
        return t;
    }

    return null;
}

The lecturer had given a simple implementation of the tree. The class is called BinaryTree and if you want to make a node link to it then you specify what the right and left child BinaryTree node are.

A node has two links (one to the left and the other to the right child) and there is no link to the head.

So with the current method I am able to successful do a preorder traversal, I tested by writing the print statements of what the element stored at the node is.

But when I run the tests, it tells me that the next preorder node from A is B, and the next preorder node from K throws a null exception but it should be I?

Any ideas of where I am going wrong? The variable prev should hold a reference to the last node visited so if it equals to node v (which is the node I specify, to return the node after v), shouldn't I get the next node?

Answer 1

Here is an implementation of how a preorder traversal is done recursively.

public void preOrderTraversal(BinaryTree root){
    if(root == null) return;

    System.out.println(root);
    preOrderTraversal(root.getLeft());
    preOrderTraversal(root.getRight());
}

Notes:

1. I am not sure of your approach; why are you returning a node? In any case, when the root is null with that approach you can return an "emptyNode" and deal with it by an if statement.
2. As you can see I am only dealing with the root, with any level the root changes. Try to visualize this with a run-through and you will understand this concept.
3. You are missing a check for null nodes at the beginning (especially for t).

You can continue to adapt your results to this.

A final note is for the run-time complexity of this approach, I'd highly recommend understanding run-time complexities for recursive functions. It will help you a lot in the future. Check this wikipedia article for recurrence relations.