Podcast Ch18d Title: Binary Search Tree Iterator Description: Additional operations first and last; the BST iterator Participants: Barry Kurtz (instructor); John Helfert and Tobie Williams (students) Textbook: Data Structures for Java; William H. Ford and William R. Topp

Additional STree Operations The method first() must return the smallest element in the tree. In a binary search tree, this element is in the leftmost node from root.

Additional STree Operations (continued) // returns the first (least) element in this // binary search tree public T first() { STNode<T> nextNode = root; // if the set is empty, return null if (nextNode == null) return null; // first node is the furthest left from root while (nextNode.left != null) nextNode = nextNode.left; return nextNode.nodeValue; }

Additional STree Operations (concluded) The method last() must return the largest element in the tree. In a binary search tree, this element is in the right-most node from root. Implement the methods by starting at the root and scan down the path of right children. The value of the last node on the path is the maximum value in the binary search tree. A binary search tree has best-case search time O(log2n), worst-case time O(n), and average time O(log2n).

Practice Coding Write the method last to find the largest item in a BST?

The STree Iterator Implement an iterative traversal of the elements using the STNode parent field. The figure illustrates the order of scan for the twenty elements in the search tree. If the current iterator position is node 18, the next two elements are found by moving up to the parent 20 and then down the right subtree to node 22 and node 23. From node 23 the next element is the root node 25. We locate the value by scanning up the path of parents four levels.

The STree Iterator (continued)

Student Question Given the tree Starting with #0, label the nodes in the order they will be visited by an iterator.

Practice Problem Given the tree Starting with #0, label the nodes in the order they will be visited by an iterator.

IteratorImpl Inner Class private class IteratorImpl implements Iterator<T> { // set expectedModCount to the number of list // changes at the time of iterator creation private int expectedModCount = modCount; // node of the last value returned by next() // if that value was deleted by the iterator // method remove() private STNode<T> lastReturned = null; // node whose value is returned a subsequent // call to next() private STNode<T> nextNode = null; ... }

IteratorImpl Inner Class (continued) The first node is the one far-left from the root. Subsequent nodes are visited in LNR order.

IteratorImpl Inner Class (cont) // constructor IteratorImpl() { nextNode = root; // if the tree is not empty, the first node // inorder is the farthest node left from root if (nextNode != null) while (nextNode.left != null) nextNode = nextNode.left; } // returns an iterator for the elements in the tree public Iterator<T> iterator() return new IteratorImpl();

Implementing next() If the right subtree is not empty, obtain the next node in order by moving to the right child and then moving left until encountering a null subtree. If the right subtree is empty, obtain the next node in order by following a chain of parent references until locating a parent, P, for which the current node, nodePtr, is a left child. Node P is the next node in order.

Implementing next() (continued) When invoking next() from the node with the tree's maximum value, the right subtree is empty. As we move up the tree, we are always a right child of our parent until we encounter the root node, whose parent is null. In this situation, the variable nextNode becomes null and we have completed an iteration.

Implementing next() (continued) // returns the next element in the iteration; // throws NoSuchElementException if iteration // has no more elements public T next() { // check that the iterator is in a consistent // state; throws ConcurrentModificationException // if it is not checkIteratorState(); // check if the iteration has another element; // if not, throw NoSuchElementException if (nextNode == null) throw new NoSuchElementException( "Iteration has no more elements"); // save current value of next in lastReturned lastReturned = nextNode; // set nextNode to the next node in order STNode<T> p;

Implementing next() (cont) // have already processed the left subtree, // and there is no right subtree; move up // the tree, looking for a parent for // which nextNode is a left child, stopping // if the parent becomes null; a non-null // parent is the successor; if parent is null, // the original node was the last node inorder p = nextNode.parent; if (nextNode.right != null) { // successor is the furthest left node of // right subtree nextNode = nextNode.right; while (nextNode.left != null) nextNode = nextNode.left; }

Implementing next() (concluded) else { while (p != null && nextNode == p.right) nextNode = p; p = p.parent; } // if we were previously at the right-most // node in the tree, nextNode = null return lastReturned.nodeValue;

Implementing hasNext() The iteration is complete when nextNode becomes null. The implementation of hasNext() simply verifies that nextNode is not equal to null. // returns true if the tree has more // unvisited elements public boolean hasNext() { // elements remain if nextNode is not null return nextNode != null; }

Student Question How would you write a non-recursive iterator if you did not have a parent pointer?

Practice Problem Write a non-recursive iterator assuming you do not have a parent pointer? The previous discussion by John and Tobie should be useful.