CHAPTER 10 Binary Search Trees Java Software Structures

CHAPTER 10: Binary Search Trees Java Software Structures: Designing and Using Data Structures Third Edition John Lewis & Joseph Chase Addison Wesley is an imprint of © 2010 Pearson Addison-Wesley. All rights reserved.

Chapter Objectives • Define a binary search tree abstract data structure • Demonstrate how a binary search tree can be used to solve problems • Examine various binary search tree implementations • Compare binary search tree implementations 1 -2 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -2

Binary Search Trees • A binary search tree is a binary tree with the added property that for each node, the left child is less than the parent is less than or equal to the right child • Given this refinement of our earlier definition of a binary tree, we can now include additional operations 1 -3 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -3

The operations on a binary search tree 1 -4 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -4

Binary. Search. Tree. ADT /** * Binary. Search. Tree. ADT defines the interface to a binary search tree. * * @author Dr. Lewis * @author Dr. Chase * @version 1. 0, 8/19/08 */ package jss 2; public interface Binary. Search. Tree. ADT<T> extends Binary. Tree. ADT<T> { /** * Adds the specified element to the proper location in this tree. * * @param element the element to be added to this tree */ public void add. Element (T element); 1 -5 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -5

Binary. Search. Tree. ADT (continued) /** * Removes and returns the specified element from this tree. * * @param target. Element the element to be removed from this tree * @return the element removed from this tree */ public T remove. Element (T target. Element); /** * Removes all occurences of the specified element from this tree. * * @param target. Element the element that the list will have all instances * of it removed */ public void remove. All. Occurrences (T target. Element); /** * Removes and returns the smallest element from this tree. * * @return the smallest element from this tree. */ public T remove. Min(); © 2010 Pearson Addison-Wesley. All rights reserved. 1 -6

Binary. Search. Tree. ADT (continued) /** * Removes and returns the largest element from this tree. * * @return the largest element from this tree */ public T remove. Max(); /** * Returns a reference to the smallest element in this tree. * * @return a reference to the smallest element in this tree */ public T find. Min(); /** * Returns a reference to the largest element in this tree. * * @return a reference to the largest element in this tree */ public T find. Max(); } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -7

UML description of the Binary. Search. Tree. ADT 1 -8 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -8

Implementing Binary Search Trees With Links • We can simply extend our definition of a Linked. Binary. Tree to create a Linked. Binary. Search. Tree • This class will provide two constructors, one to create an empty tree and the other to create a one-element binary tree 1 -9 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -9

Linked. Binary. Search. Tree /** * Linked. Binary. Search. Tree implements the Binary. Search. Tree. ADT interface * with links. * * @author Dr. Lewis * @author Dr. Chase * @version 1. 0, 8/19/08 */ package jss 2; import jss 2. exceptions. *; import jss 2. *; public class Linked. Binary. Search. Tree<T> extends Linked. Binary. Tree<T> implements Binary. Search. Tree. ADT<T> { /** * Creates an empty binary search tree. */ public Linked. Binary. Search. Tree() { super(); } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -10

Linked. Binary. Search. Tree (cont. ) /** * Creates a binary search with the specified element as its root. * * @param element the element that will be the root of the new binary * search tree */ public Linked. Binary. Search. Tree (T element) { super (element); } 1 -11 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -11

Implementing Binary Search Trees With Links • Now that we know more about how this tree is to be used (and structured) it is possible to define a method to add an element to the tree • The add. Element method finds the proper location for the given element and adds it there as a leaf 1 -12 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -12

Linked. Binary. Search. Tree add. Element /** * Adds the specified object to the binary search tree in the * appropriate position according to its key value. Note that * equal elements are added to the right. * * @param element the element to be added to the binary search tree */ public void add. Element (T element) { Binary. Tree. Node<T> temp = new Binary. Tree. Node<T> (element); Comparable<T> comparable. Element = (Comparable<T>)element; if (is. Empty()) root = temp; else { Binary. Tree. Node<T> current = root; boolean added = false; while (!added) { if (comparable. Element. compare. To(current. element) < 0) { if (current. left == null) © 2010 Pearson Addison-Wesley. All rights reserved. 1 -13

Linked. Binary. Search. Tree – add. Element (continued) { current. left = temp; added = true; } else current = current. left; } else { if (current. right == null) { current. right = temp; added = true; } else current = current. right; } } } count++; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -14

Adding elements to a binary search tree 1 -15 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -15

Removing Elements • Removing elements from a binary search tree requires – Finding the element to be removed – If that element is not a leaf, then replace it with its inorder successor – Return the removed element 1 -16 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -16

Removing Elements • The remove. Element method makes use of a private replacement method to find the proper element to replace a non-leaf element that is removed 1 -17 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -17

Linked. Binary. Search. Tree remove. Element /** * Removes the first element that matches the specified target * element from the binary search tree and returns a reference to * it. Throws a Element. Not. Found. Exception if the specified target * element is not found in the binary search tree. * * @param target. Element the element being sought in the binary * search tree * @throws Element. Not. Found. Exception if an element not found exception occurs */ public T remove. Element (T target. Element) throws Element. Not. Found. Exception { T result = null; if (!is. Empty()) { if (((Comparable)target. Element). equals(root. element)) { result = root. element; root = replacement (root); count--; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -18

Linked. Binary. Search. Tree – remove. Element (continued) else { Binary. Tree. Node<T> current, parent = root; boolean found = false; if (((Comparable)target. Element). compare. To(root. element) < 0) current = root. left; else current = root. right; while (current != null && !found) { if (target. Element. equals(current. element)) { found = true; count--; result = current. element; if (current == parent. left) { parent. left = replacement (current); } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -19

Linked. Binary. Search. Tree – remove. Element (continued) else { parent. right = replacement (current); } } else { parent = current; if (((Comparable)target. Element). compare. To(current. element) < 0) current = current. left; else current = current. right; } } //while if (!found) throw new Element. Not. Found. Exception("binary search tree"); } } // end outer if return result; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -20

Linked. Binary. Search. Tree replacement /** * Returns a reference to a node that will replace the one * specified for removal. In the case where the removed node has * two children, the inorder successor is used as its replacement. * * @param node the node to be removeed * @return a reference to the replacing node */ protected Binary. Tree. Node<T> replacement (Binary. Tree. Node<T> node) { Binary. Tree. Node<T> result = null; if ((node. left == null)&&(node. right==null)) result = null; else if ((node. left != null)&&(node. right==null)) result = node. left; else if ((node. left == null)&&(node. right != null)) result = node. right; 1 -21 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -21

Linked. Binary. Search. Tree – replacement (continued) else { Binary. Tree. Node<T> current = node. right; Binary. Tree. Node<T> parent = node; while (current. left != null) { parent = current; current = current. left; } if (node. right == current) current. left = node. left; else { parent. left = current. right; current. right = node. right; current. left = node. left; } result = current; } return result; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -22

Removing elements from a binary tree 1 -23 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -23

Removing All Occurrences • The remove. All. Occurrences method removes all occurrences of an element from the tree • This method uses the remove. Element method • This method makes a distinction between the first call and successive calls to the remove. Element method 1 -24 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -24

Linked. Binary. Search. Tree remove. All. Occurrences /** * Removes elements that match the specified target element from * the binary search tree. Throws a Element. Not. Found. Exception if * the sepcified target element is not found in this tree. * * @param target. Element the element being sought in the binary * search tree * @throws Element. Not. Found. Exception if an element not found exception occurs */ public void remove. All. Occurrences (T target. Element) throws Element. Not. Found. Exception { remove. Element(target. Element); try { while (contains( (T) target. Element)) remove. Element(target. Element); } catch (Exception Element. Not. Found. Exception) { } } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -25

The remove. Min Operation • There are three cases for the location of the minimum element in a binary search tree: – If the root has no left child, then the root is the minimum element and the right child of the root becomes the new root – If the leftmost node of the tree is a leaf, then we set its parent’s left child reference to null – If the leftmost node of the tree is an internal node, then we set its parent’s left child reference to point to the right child of the node to be removed 1 -26 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -26

Linked. Binary. Search. Tree remove. Min /** * Removes the node with the least value from the binary search * tree and returns a reference to its element. Throws an * Empty. Binary. Search. Tree. Exception if this tree is empty. * * @return a reference to the node with the least * value * @throws Empty. Collection. Exception if an empty collection exception occurs */ public T remove. Min() throws Empty. Collection. Exception { T result = null; if (is. Empty()) throw new Empty. Collection. Exception ("binary search tree"); else { if (root. left == null) { result = root. element; root = root. right; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -27

Linked. Binary. Search. Tree – remove. Min (continued) else { Binary. Tree. Node<T> parent = root; Binary. Tree. Node<T> current = root. left; while (current. left != null) { parent = current; current = current. left; } result = current. element; parent. left = current. right; } count--; } return result; } 1 -28 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -28

Removing the minimum element from a binary search tree 1 -29 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -29

Implementing Binary Search Trees with Arrays • Just as we have done with the Linked. Binary. Search. Tree we can extend our Array. Binary. Tree from chapter 9 to create an Array. Binary. Search. Tree 1 -30 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -30

Array. Binary. Search. Tree /** * Array. Binary. Search. Tree implements a binary search tree using an array. * * @author Dr. Lewis * @author Dr. Chase * @version 1. 0, 8/19/08 */ package jss 2; import java. util. Iterator; import jss 2. exceptions. *; import jss 2. *; public class Array. Binary. Search. Tree<T> extends Array. Binary. Tree<T> implements Binary. Search. Tree. ADT<T> { protected int height; protected int max. Index; 1 -31 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -31

Array. Binary. Search. Tree (continued) /** * Creates an empty binary search tree. */ public Array. Binary. Search. Tree() { super(); height = 0; max. Index = -1; } /** * Creates a binary search with the specified element as its * root. * * @param element the element that will become the root of the new tree */ public Array. Binary. Search. Tree (T element) { super(element); height = 1; max. Index = 0; } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -32

Array. Binary. Search. Tree - add. Element /** * Adds the specified object to this binary search tree in the * appropriate position according to its key value. Note that * equal elements are added to the right. Also note that the * index of the left child of the current index can be found by * doubling the current index and adding 1. Finding the index * of the right child can be calculated by doubling the current * index and adding 2. * * @param element the element to be added to the search tree */ public void add. Element (T element) { if (tree. length < max. Index*2+3) expand. Capacity(); Comparable<T> tempelement = (Comparable<T>)element; if (is. Empty()) { tree[0] = element; max. Index = 0; } else © 2010 Pearson Addison-Wesley. All rights reserved. 1 -33

Array. Binary. Search. Tree - add. Element { boolean added = false; int current. Index = 0; while (!added) { if (tempelement. compare. To((tree[current. Index]) ) < 0) { /** go left */ if (tree[current. Index*2+1] == null) { tree[current. Index*2+1] = element; added = true; if (current. Index*2+1 > max. Index) max. Index = current. Index*2+1; } else current. Index = current. Index*2+1; } 1 -34 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -34

Array. Binary. Search. Tree - add. Element else { /** go right */ if (tree[current. Index*2+2] == null) { tree[current. Index*2+2] = element; added = true; if (current. Index*2+2 > max. Index) max. Index = current. Index*2+2; } else current. Index = current. Index*2+2; } } } height = (int)(Math. log(max. Index + 1) / Math. log(2)) + 1; count++; } 1 -35 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -35

Using Binary Search Trees: Implementing Ordered Lists • Lets look at an example using a binary search tree to provide an efficient implementation of an ordered list • For simplicity, we will implement both the List. ADT and the Ordered. List. ADT in the Binary. Search. Tree. List class 1 -36 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -36

The common operations on a list 1 -37 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -37

The operation particular to an ordered list 1 -38 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -38

The Binary. Search. Tree. List class /** * Binary. Search. Tree. List represents an ordered list implemented using a binary * search tree. * * @author Dr. Lewis * @author Dr. Chase * @version 1. 0, 8/19/08 */ package jss 2; import jss 2. exceptions. *; import java. util. Iterator; public class Binary. Search. Tree. List<T> extends Linked. Binary. Search. Tree<T> implements List. ADT<T>, Ordered. List. ADT<T>, Iterable<T> { /** * Creates an empty Binary. Search. Tree. List. */ public Binary. Search. Tree. List() { super(); } © 2010 Pearson Addison-Wesley. All rights reserved. 1 -39

The Binary. Search. Tree. List class (cont. ) /** * Adds the given element to this list. * * @param element the element to be added to this list */ public void add (T element) { add. Element(element); } /** * Removes and returns the first element from this list. * * @return the first element in this list */ public T remove. First () { return remove. Min(); } 1 -40 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -40

The Binary. Search. Tree. List class (cont. ) /** * Removes and returns the last element from this list. * * @return the last element from this list */ public T remove. Last () { return remove. Max(); } /** * Removes and returns the specified element from this list. * * @param element the element being sought in this list * @return the element from the list that matches the target */ public T remove (T element) { return remove. Element(element); } 1 -41 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -41

The Binary. Search. Tree. List class (cont. ) /** * Returns a reference to the first element on this list. * * @return a reference to the first element in this list */ public T first () { return find. Min(); } /** * Returns a reference to the last element on this list. * * @return a reference to the last element in this list */ public T last () { return find. Max(); } 1 -42 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -42

The Binary. Search. Tree. List class (cont. ) /** * Returns an iterator for the list. * * @return an iterator over the elements in this list */ public Iterator<T> iterator() { return iterator. In. Order(); } } 1 -43 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -43

Analysis of linked list and binary search tree implementations of an ordered list 1 -44 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -44

Balanced Binary Search Trees • Why is our balance assumption so important? • Lets look at what happens if we insert the following numbers in order without rebalancing the tree: 3 5 9 12 18 20 1 -45 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -45

A degenerate binary tree 1 -46 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -46

Degenerate Binary Trees • The resulting tree is called a degenerate binary tree • Degenerate binary search trees are far less efficient than balanced binary search trees (O(n) on find as opposed to O(logn)) 1 -47 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -47

Balancing Binary Trees • There are many approaches to balancing binary trees • One method is brute force – Write an inorder traversal to a file – Use a recursive binary search of the file to rebuild the tree 1 -48 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -48

Balancing Binary Trees • Better solutions involve algorithms such as red-black trees and AVL trees that persistently maintain the balance of the tree • Most all of these algorithms make use of rotations to balance the tree • Lets examine each of the possible rotations 1 -49 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -49

Right Rotation • Right rotation will solve an imbalance if it is caused by a long path in the left sub-tree of the left child of the root – Make the left child element of the root the new root element. – Make the former root element the right child element of the new root. – Make the right child of what was the left child of the former root the new left child of the former root. 1 -50 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -50

Unbalanced tree and balanced tree after a right rotation 1 -51 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -51

Left Rotation • Left rotation will solve an imbalance if it is caused by a long path in the right sub-tree of the right child of the root – Make the right child element of the root the new root element. – Make the former root element the left child element of the new root. – Make the left child of what was the right child of the former root the new right child of the former root. © 2010 Pearson Addison-Wesley. All rights reserved. 1 -52

Unbalanced tree and balanced tree after a left rotation 1 -53 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -53

Rightleft Rotation • Rightleft rotation will solve an imbalance if it is caused by a long path in the left sub-tree of the right child of the root • Perform a right rotation of the left child of the right child of the root around the right child of the root, and then perform a left rotation of the resulting right child of the root around the root 1 -54 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -54

A rightleft rotation 1 -55 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -55

Leftright Rotation • Leftright rotation will solve an imbalance if it is caused by a long path in the right sub-tree of the left child of the root • Perform a left rotation of the right child of the left child of the root around the left child of the root, and then perform a right rotation of the resulting left child of the root around the root 1 -56 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -56

A leftright rotation 1 -57 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -57

AVL Trees • AVL trees keep track of the difference in height between the right and left sub-trees for each node • This difference is called the balance factor • If the balance factor of any node is less than -1 or greater than 1, then that sub-tree needs to be rebalanced • The balance factor of any node can only be changed through either insertion or deletion of nodes in the tree 1 -58 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -58

AVL Trees • If the balance factor of a node is -2, this means the left sub-tree has a path that is too long • If the balance factor of the left child is -1, this means that the long path is the left sub-tree of the left child • In this case, a simple right rotation of the left child around the original node will solve the imbalance 1 -59 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -59

A right rotation in an AVL tree 1 -60 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -60

AVL Trees • If the balance factor of a node is +2, this means the right sub-tree has a path that is too long • Then if the balance factor of the right child is +1, this means that the long path is the right sub-tree of the right child • In this case, a simple left rotation of the right child around the original node will solve the imbalance 1 -61 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -61

AVL Trees • If the balance factor of a node is +2, this means the right sub-tree has a path that is too long • Then if the balance factor of the right child is -1, this means that the long path is the left sub-tree of the right child • In this case, a rightleft double rotation will solve the imbalance 1 -62 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -62

A rightleft rotation in an AVL tree 1 -63 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -63

AVL Trees • If the balance factor of a node is -2, this means the left sub-tree has a path that is too long • Then if the balance factor of the left child is +1, this means that the long path is the right sub-tree of the left child • In this case, a leftright double rotation will solve the imbalance 1 -64 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -64

Red/Black Trees • Red/Black trees provide another alternative implementation of balanced binary search trees • A red/black tree is a balanced binary search tree where each node stores a color (usually implemented as a boolean) • The following rules govern the color of a node: – The root is black – All children of a red node are black – Every path from the root to a leaf contains the same number of black nodes 1 -65 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -65

Valid red/black trees 1 -66 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -66

Insertion into Red/Black Trees • The color of a new element is set to red • Once the new element has been inserted, the tree is rebalanced/recolored as needed to to maintain the properties of a red/black tree • This process is iterative beginning at the point of insertion and working up the tree toward the root • The process terminates when we reach the root or when the parent of the current element is black 1 -67 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -67

Insertion into Red/Black Trees • In each iteration of the rebalancing process, we will focus on the color of the sibling of the parent of the current node • There are two possibilities for the parent of the current node: – The parent could be a left child – The parent could be a right child • The color of a null node is considered to be black 1 -68 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -68

Insertion into Red/Black Trees • In the case where the parent of the current node is a right child, there are two cases – parentsleftsibling. color == red – parentsleftsibling. color == black • If parentsleftsibling. color is red then the processing steps are: – – Set the color of current’s parent to black. Set the color of parentsleftsibling to black. Set the color of current’s grandparent to red. Set current to point to the grandparent of current 1 -69 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -69

Red/black tree after insertion 1 -70 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -70

Insertion into Red/Black Trees • If parentsleftsibling. color is black, we first must check to see if current is a left child or a right child • If current is is a left child, then we must set current equal to current. parent and then rotate current. left to the right • The we continue as if current were a right child to begin with: – Set the color of current’s parent to black. – Set the color of current’s grandparent to red. – If current’s grandparent does not equal null, then rotate current’s parent to the left around current’s grandparent 1 -71 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -71

Insertion into Red/Black Trees • In the case where the parent of current is a left child, there are two cases: either parentsrightsibling. color == red or parentsrightsibling. color == black • If parentsrightsibling. color == red then the processing steps are: – – Set the color of current’s parent to black. Set the color of parentsrightsibling to black. Set the color of current’s grandparent to red. Set current to point to the grandparent of current 1 -72 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -72

Red/black tree after insertion 1 -73 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -73

Insertion into Red/Black Trees • If parentsrightsibling. color == black then we first need to check to see if current is a left or right child • If current is a right child then we set current equal to current. parent then rotate current. right to the left around current • We then continue as if current was left child to begin with: – Set the color of current’s parent to black. – Set the color of current’s grandparent to red. – If current’s grandparent does not equal null, then rotate current’s parent to the right around current’s grandparent 1 -74 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -74

Element Removal from Red/Black Trees • As with insertion, the tree will need to be rebalanced/recolored after the removal of an element • Again, the process is an iterative one beginning at the point of removal and continuing up the tree toward the root • This process terminates when we reach the root or when current. color == red • Like insertion, the cases for removal are symmetrical depending upon whether current is a left or right child • In insertion, we focused on the color of the sibling of the parent • In removal, we focus on the color of the sibling of current keeping in mind that a null node is considered to be black 1 -75 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -75

Element Removal from Red/Black Trees • We will only examine the cases where current is a right child, the other cases are easily derived • If the sibling’s color is red then we begin with the following steps: – – Set the color of the sibling to black. Set the color of current’s parent to red. Rotate the sibling right around current’s parent. Set the sibling equal to the left child of current’s parent 1 -76 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -76

Red/black tree after removal 1 -77 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -77

Element Removal from Red/Black Trees • Next our processing continues regardless of whether the original sibling was red or black • Now our processing is divided into two cases based upon the color of the children of the sibling • If both of the children of the sibling are black then: – Set the color of the sibling to red. – Set current equal to current’s parent 1 -78 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -78

Element Removal from Red/Black Trees • If the children of the sibling are not both black, then we check to see if the left child of the sibling is black • If it is, then we must complete the following steps before continuing: – – Set the color of the sibling’s right child to black. Set the color of the sibling to red. Rotate the sibling’s right child left around the sibling. Set the sibling equal to the left child of current’s parent 1 -79 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -79

Element Removal from Red/Black Trees • Then to complete the process in the case when both of the children of the sibling are not black: – – – Set the color of the sibling to the color of current’s parent. Set the color of current’s parent to black. Set the color of the sibling’s left child to black. Rotate the sibling right around current’s parent. Set current equal to the root 1 -80 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -80

Binary Search Trees in the Java Collections API • Java provides two implementations of balanced binary search trees – Tree. Set – Tree. Map • In order to understand the difference between these two, we must first discuss the difference between a set and a map 1 -81 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -81

Sets and Maps • In the terminology of the Java Collections API, all of the collections we have discussed thus far would be considered sets • A set is a collection where all of the data associated with an object is stored in the collection • A map is a collection where keys that reference an object are stored in the collection but the remaining data is stored separately 1 -82 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -82

Sets and Maps • Maps are useful because they allow us to manipulate keys within a collection rather than the entire object • This allows collections to be smaller, more efficient, and easier to manage • This also allows for the same object to be part of multiple collections by having keys in each 1 -83 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -83

The Tree. Set and Tree. Map Classes • Both the Tree. Set and Tree. Map classes are red/black tree implementations of a balanced binary search tree 1 -84 © 2010 Pearson Addison-Wesley. All rights reserved. 1 -84