Binary Trees Like a list, a (binary) tree can be empty or non-empty. In class we will explore a state-based implementation, similar to the LRStruct You are expected to read about alternate implementation strategies in the textbook

Methods of a Binary Recursive Structure (BRStruct) method to grow tree: insertRoot method to shrink tree: removeRoot accessor/mutator pair for datum: setDatum/getDatum accessor/mutator pair for left: setLeft/getLeft accessor/mutator pair for right: setRight/getRight visitor support: execute

State transitions Empty  NonEmpty: insertRoot on an empty tree NonEmpty  Empty: removeRoot from a tree that is a leaf –A leaf is a tree both of whose children are empty. No other state transitions can occur.

Moreover… insertRoot throws an exception if called on a nonEmpty tree removeRoot throws an exception if called on a non-leaf tree Isn’t this very restrictive?

Yes…but It is very restrictive, but for good reasons: –What would it mean to add a root to a tree that already has one? –If you could remove the root from a tree with nonEmpty children, what would become of those children?

Usage of a BRS A BRStruct is not used “raw”, but is used to implement more specialized trees. Consider how we defined the SortedList in terms of the LRStruct: by composition. We will now explore how to define a sorted binary tree (a Binary Search Tree) by composition with a BRStruct.

Binary Search Tree (BST) A binary search tree (BST) is a binary tree which maintains its elements in order. The BST order condition requires that, for every non-empty tree, the value at the root is greater than every value in the tree’s left subtree, and less than every value in the tree’s right subtree.

Example 1 Does this tree satisfy the BST order condition? Fred WilmaBetty BarneyPebbles

No! Barney Fred < Wilma Fred WilmaBetty BarneyPebbles

Example 2 Does this tree satisfy the BST order condition? Fred WilmaBetty BarneyPebbles

Yes! Barney < Betty < Fred < Pebbles < Wilma Fred WilmaBetty BarneyPebbles

Example 3 Does this tree satisfy the BST order condition? Fred WilmaBetty BarneyPebbles

No! Betty > Barney < Fred < Pebbles < Wilma Fred WilmaBetty BarneyPebbles

Example 4 …but if we swap Betty & Barney, we restore order! Fred Wilma Betty Barney Pebbles

Operations on a BST public BST insert(E item) public BST remove(E item) public boolean member(E item) How do we support these? They don’t exist as methods on the BRStruct.

Visitors to the rescue! Visitors on the BRStruct have the same basic structure as on the LRStruct: they deal with two cases: emptyCase and nonEmpty Case. In our example: public BRStruct emptyCase(BRStruct host, E arg) public BRStruct nonEmptyCase(BRStruct host, E arg)

Example: toStringVisitor public class ToStringVisitor extends IAlgo { public static final ToStringVisitor SINGLETON = new ToStringVisitor(); private ToStringVisitor() {} public Object emptyCase(BRS host, Object input) { return "[]"; } public Object nonEmptyCase(BRS host, Object input) { return "[" + host.getLeft().execute(this, null) + " " + host.getRoot().toString() + " " + host.getRight().execute(this, null) + "]"; }

Back to the BST Let’s first consider the insert operation. We must consider two possibilities: –the underlying BRStruct is empty just insertRoot –the underlying BRStruct is nonEmpty we can’t insertRoot, because the BRStruct is nonEmpty compare new item with root – determine whether new item belongs in left or right subtree – insert recursively into correct subtree EXERCISE: DEFINE THIS VISITOR

A first cut at the visitor public class InsertVisitor implements IAlgo { public Object emptyCase(BRStruct host, E item) { return host.insertRoot(item); } public Object nonEmptyCase(BRStruct host, E item) { if (item.compareTo(host.getDatum()) < 0 ) { // item belongs in left subtree return host.getLeft().execute(this, item); } else if (item.compareTo(host.getRoot()) > 0 ) { // item belongs in right subtree return host.getRight().execute(this, item); } else { // item is already in tree (Note UNIQUENESS ASSUMPTION) return host; }

How about determining membership for an item? We must consider two possibilities: –the underlying BRStruct is empty just item was not found –the underlying BRStruct is nonEmpty compare new item with root – determine whether new item has been found, or whether it would be in left or right subtree – look recursively into correct subtree EXERCISE: DEFINE THIS VISITOR

A first cut at the visitor public class MemberVisitor implements IAlgo { public Boolean emptyCase(BRStruct host, E item) { return false; } public Boolean nonEmptyCase(BRStruct host, E item) { if (item.compareTo(host.getDatum()) < 0 ) { // item belongs in left subtree return host.getLeft().execute(this, item); } else if (item.compareTo(host.getRoot()) > 0 ) { // item belongs in right subtree return host.getRight().execute(this, item); } else { // item is in tree return true; }

Did you notice similarity? The structure of the insert and membership visitors was the same. A small number of details differed. Let’s unify! EXERCISE: DEFINE THIS VISITOR

The find visitor public class MemberVisitor implements IAlgo > { public BRStruct emptyCase(BRStruct host, E item) { // item is in not in tree, but would belong in host return host; } public BRStruct nonEmptyCase(BRStruct host, E item) { if (item.compareTo(host.getDatum()) < 0 ) { // item belongs in left subtree return host.getLeft().execute(this, item); } else if (item.compareTo(host.getRoot()) > 0 ) { // item belongs in right subtree return host.getRight().execute(this, item); } else { // item appears in tree as datum of host return host; }

Look at BST implementation insert, remove and member ALL make use of the same FindVisitor: –first find the insertionPoint, –then do the right thing.

The insert method public BST insert(E item) { BRStruct tree = _tree.execute(new FindVisitor (), item); only insert item into insertPoint if empty (duplicates ignored) return this; } EXERCISE: SOLVE PURPLE PROBLEM

The insert method public BST insert(E item) { BRStruct tree = _tree.execute(new FindVisitor (), item); tree.execute(new IAlgo () { public Object emptyCase(BRStruct host, E input) { host.insertRoot(input); return null; } public Object nonEmptyCase(BRStruct host, E input) { return null; } }, item); return this; }

The member method public BST member(E item) { BRStruct tree = _tree.execute(new FindVisitor (), item); return whether insertPoint is empty or nonEmpty } EXERCISE: SOLVE PURPLE PROBLEM

The member method public BST member(E item) { BRStruct tree = _tree.execute(new FindVisitor (), item); return tree.execute(new IAlgo () { public Boolean emptyCase(BRS host, Object _) { return false; } public Boolean nonEmptyCase(BRS host, Object _) { return true; } }, null); }

The remove method? Next time!