Question

For instance, if I had

  A
 / \
 B  C
/  
D 

I would want the next addition to be:

  A
 / \
 B  C
/ \ 
D  E

But I’m having a lot of trouble detecting where the next spot for the item to input will be. I have the following code:

    public static BinaryTree<String> addToTree(BinaryTree<String> tree, String name) {
        if (tree.getLeft() == null) {
            BinaryTree<String> newTree = new BinaryTree<String>();
            newTree.makeRoot(name);
            tree.attachLeft(newTree);
        }
        else if (tree.getRight() == null) {
            BinaryTree<String> newTree = new BinaryTree<String>();
            newTree.makeRoot(name);
            tree.attachRight(newTree);
        }
        // Both are non-null
        else {
            if (tree.getLeft().getLeft() == null || tree.getLeft().getRight() == null) {
                tree.attachLeft(addToTree(tree.getLeft(), name));
            }
            else if (tree.getRight().getLeft() == null || tree.getRight().getRight() == null) {
                tree.attachRight(addToTree(tree.getRight(), name));
            }
        }

        return tree;
    }

But it will only work for up to a three level tree. If I try to add the fourth, it no longer adds any.

How do I implement it so it will figure out where the next item is null, and then add it there?

I also thought of having a checkNullity() method, wherein I’d take a tree, and check if its children were null, but I was also having trouble figuring out how to get the children’s children. I wanted to find where it was null and then add it there.

Could anyone offer some input?

Answer 1

You can modify breadth first traversal to accomplish this I think. When you pop up the items from the queue, check if any of the children is empty. The first empty child slot is the place you want to add to.

addNode(root, newNode) 
  q = empty queue
  q.enqueue(root)
  while not q.empty do
    node := q.dequeue()
    if node.left == null
      //create new node as nodes left child
      return
    q.enqueue(node.left)
    if node.right == null
      //create new node as nodes right child
      return
    q.enqueue(node.right)