Binary Search Trees (BSTs) Implementing the BST ADT. Class #16: Binary Search Trees 1, 3, 5, 7, 8, 9, 11, 12, 13, 15!
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1 Class #16: Binary Search Trees Software Design III (CS 40): M. Allen, 29 February 2016 Binary Search Trees (BSTs)! A BST is a binary tree that stores its data in an efficiently searchable form! Each node has a key-value associated with it! May or may not be identical to the data being stored! Every node has a key greater than every key in left-hand subtree, and less than every key in right-hand subtree! Every key is thus unique 1 The idea can be extended to trees that allow duplicate elements/keys, although some of the implementation will be a bit more involved. Monday, 29 Feb. 2016" Software Design III (CS 40)" 2" Binary Search Trees (BSTs)! Since the tree stores its data always in a sorted order, we can easily generate that ordering by traversing the tree! A simple in-order traversal (left, parent, right) produces the ascending sorted order 1 1,,, 7,, 9, 11,,,! Implementing the BST ADT! We create the tree using a linking-node structure, using a simple extension of the idea behind a linked list! For simplicity, we treat the element itself as the key-value for the node that contains it private static class BinaryNode<T>! private T element;! private BinaryNode<T> left;! private BinaryNode<T> right;! private BinaryNode( T elt, BinaryNode<T> lt, BinaryNode<T> rt )! element = elt;! left = lt;! right = rt;! Monday, 29 Feb. 2016" Software Design III (CS 40)" " Monday, 29 Feb. 2016" Software Design III (CS 40)" 4" 1
2 Implementing the BST ADT! The tree is parameterized with a Comparable type, so that elements can be treated as key-values and ordered properly class BinarySearchTree<T extends Comparable<? super T>>!! private BinaryNode<T> root;! public BinarySearchTree( )! root = null;! /** Methods go here... */! This type parameter guarantees that the object data can be compared using the compareto() method. This method can be contained in class T itself, or it can be inherited from an ancestor class (super). Inserting into a BST! To keep the tree properly sorted, insertions start at the root and move down to the proper place! At each node, we will go left or right based on return value ( n, 0, +m) of the compareto() method! To insert: 6 6 < 6 > < 7 Monday, 29 Feb. 2016" Software Design III (CS 40)" " Monday, 29 Feb. 2016" Software Design III (CS 40)" 6" Inserting into a BST public void insert( T x )! root = insert( x, root );! private BinaryNode<T> insert( T x, BinaryNode<T> t )! if ( t == null )! return new BinaryNode<T>( x, null, null );! int compareresult = x.compareto( t.element );! if ( compareresult < 0 )! t.left = insert( x, t.left );! else if ( compareresult > 0 )! t.right = insert( x, t.right );! // If here, then x is a duplicate item;! // simply return original tree! Monday, 29 Feb. 2016" Software Design III (CS 40)" 7" 1 To insert: Optionally: if we allow duplicates, then insert either on left/right Searching in a BST public boolean contains( T x ) return contains( x, root ); } private boolean contains( T x, BinaryNode<T> t ) if ( t == null ) return false; int compareresult = x.compareto( t.element ); if ( compareresult < 0 ) return contains( x, t.left ); else if ( compareresult > 0 ) return contains( x, t.right ); else return true; On any non-matching key, we recursively repeat process on relevant sub-tree.! As with insertion, finding an element is a simple matter of following branches according to the output of compareto(). Monday, 29 Feb. 2016" Software Design III (CS 40)" " 2
3 Searching in a BST private BinaryNode<T> findmin( BinaryNode<T> t ) if ( t == null ) return null; else if ( t.left == null ) return t; } return findmin( t.left ); private BinaryNode<T> findmax( BinaryNode<T> t ) if ( t!= null ) while ( t.right!= null ) t = t.right; return t; Traverse to the left, recursively. Traverse to the right, iteratively.! We can also elegantly find the minimum or maximum element in a BST, simply by repeatedly going left/right (either recursively or iteratively), until we hit a leaf node Monday, 29 Feb. 2016" Software Design III (CS 40)" 9"! Deleting nodes is more complex, due to the fact that nodes may have children that must be kept in the tree! When we have no children (i.e., we are at a leaf) we can simply set the parent s node reference null! When we have only one child, we simply bypass it, as we would in a linked list! Garbage collection takes care of the actual node deletions if necessary later on Monday, 29 Feb. 2016" Software Design III (CS 40)" 10" 1 7! When we have two children, things get more complex! In order to keep the sorted order, we need to: 1. Replace the key-value for the deleted node with the smallest value below it from the right-hand sub-tree 2. Then delete that smallestvalued node (which always only has at most one child) Step 1 / 6 11 Step 2 public void remove( T x )! root = remove( x, root );! private BinaryNode<T> remove( T x, BinaryNode<T> t ) if( t == null )! int comp = x.compareto( t.element );! if( comp < 0 )! t.left = remove( x, t.left );! else if( comp > 0 )! t.right = remove( x, t.right );! else if( t.left!= null && t.right!= null )! t.element = findmin( t.right ).element;! t.right = remove( t.element, t.right );! else if( t.left!= null )! t = t.left;! else! t = t.right;! return t;! Step 1 / 6 11 Step 2 Monday, 29 Feb. 2016" Software Design III (CS 40)" 11" Monday, 29 Feb. 2016" Software Design III (CS 40)" "
4 Binary Search Tree Complexity! Runtimes of main methods depend upon the tree s depth! Depth depends upon how balanced it is! Balance depends upon the order in which things are inserted! If the tree is complete, we have the most balanced case possible, and then we can get better times Method! Worst Case! Balanced Case! boolean insert( T val )! O(n)! O(log n)! boolean remove( T val )! O(n)! O(log n)! boolean contains( T val )! O(n)! O(log n)! T findmin()! O(n)! O(log n)! T findmax()! O(n)! O(log n)! Balance in Binary Search Trees BinaryNode<T> remove( T x, BinaryNode<T> t )!! if( t == null )! int comp = x.compareto( t.element );! if( comp < 0 )! t.left = remove( x, t.left );! else if( comp > 0 )! t.right = remove( x, t.right );! else if( t.left!= null && t.right!= null )! t.element = findmin( t.right ).element;! t.right = remove( t.element, t.right );! else if( t.left!= null )! t = t.left;! else! t = t.right;! return t;!! Analysis shows average depth of a node in a randomly generated BST with n inserted nodes is O(log n)! However, trees are prone to become unbalanced! When deleting internal nodes, since we always take from right-hand side when we need to replace a 2-child node, trees tend to become left-heavy Monday, 29 Feb. 2016" Software Design III (CS 40)" " Monday, 29 Feb. 2016" Software Design III (CS 40)" 14" Balance in BSTs! It has been proven that n nodes, if we do Θ(n 2 ) random delete/insert pairs, the expected average depth will grow from O(log n) to Θ( n)! After that, we will see somewhat unpredictable fluctuations! Switching the strategy to randomly choose left/right seems to help, but this has never been definitively proven to work Adelson-Velskii & Landis (AVL) Trees! A simple self-adjusting tree with the properties: 1. It is a properly ordered BST 2. At each node, height of right/left subtrees differ by one at most AVL BST Non-AVL BST Monday, 29 Feb. 2016" Software Design III (CS 40)" " Monday, 29 Feb. 2016" Software Design III (CS 40)" 16" 4
5 Size and Efficiency of AVL Trees! As long as the balance property is maintained, it is easy to show that for any height h, the minimum number of nodes in an AVL tree of that size, N(h), is given by:! This allows us to show that: N(h) = N(h 1) + N(h 2) + 1 Where: N(0) = 1 N(1) = 2 1. For n = N(h) nodes, height of AVL tree is h(n) = O(log n) 2. Every set of m operations takes time O(m log n) This Week! Topic: Binary Search Trees and AVL Trees! Meetings: In class all week (no lab)! Read: Text, First half of chapter 04! Midterm 01: Wednesday, 02 March! Covers all material up to (and including) Stacks & Queues! Chapters 1, 2, of text! Lectures 1 9! Practice exam and key on D2L now! Office Hours: Wing 210! Tuesday & Thursday: 10:00 11:0 AM! Tuesday & Thursday: 4:00 :0 PM Monday, 29 Feb. 2016" Software Design III (CS 40)" 17" Monday, 29 Feb. 2016" Software Design III (CS 40)" 1"
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