javaavl-treetree-balancing

AVL Tree Balancing


I am working on an assignment that asks me to implement an AVL tree. I'm pretty sure I have the rotation methods correct, but I'm having trouble figuring out when to use them.

For example, the explanation in the book says that I should climb up the same path I went down to insert the node/element. However, I can't have any parent pointers.

Latest code:

public BinaryNode<T> insert(BinaryNode<T> node) {
    if (this.getElement().compareTo(node.getElement()) > 0) {
        if (this.getLeftChild() != null) {
            BinaryNode<T> b = this.getLeftChild().insert(node);

            if(!this.isBalanced()) {
                this.balance();
            }

            return b;
        } else {
            this.setLeftChild(node);
        }

    } else if (this.getElement().compareTo(node.getElement()) < 0) {
        if (this.getRightChild() != null) {
            return this.getRightChild().insert(node);
        } else {
            this.setRightChild(node);
        }
    }

    return this;
}

What I want to do here is climb back up the tree, but it can only check the balancing AFTER it inserts the node. Hence, this being in the else clause.

I also tried putting the balance code where R Samuel Klatchko suggested, but checked the balance on each insert. For example: If one inserts 7, 9, 5, 3, and 1 consecutively, I get a null pointer exception when trying to insert 1.

EDIT: One reason for the above may have something to do with the way I was doing the height. It works fine with a single right rotation if I calculate the height every time with height() but that breaks the O(log(n)) time of an AVL Tree.

Any thoughts on how to accomplish this?


Solution

  • You code is climbing up the same path you went down. Consider this code:

    if (this.getLeftChild() != null) {
        return this.getLeftChild().insert(node);
    } 
    

    and modify it slightly:

    if (this.getLeftChild() != null) {
        boolean b = this.getLeftChild().insert(node);
        // do something here
        return b;
    } 
    

    As the code returns from the recursive calls, each return brings you back to the parent. By not immediately returning the value of the recursive call, you have a chance to do your rebalancing.

    Update for latest code

    Don't forget to rebalance when you've inserted to the right.