Ch13 Binary Search Tree

Chapter 13 Binary Search Trees

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

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

FIGURE 13.1 The operations on a binary search tree

FIGURE 13.2 UML description of the BinarySearchTreeADT

Implementing Binary Search Trees With Links We can simply extend our definition of a LinkedBinaryTree to create a LinkedBinarySearchTree This class will provide two constructors, one to create an empty tree and the other to create a one-element binary tree

LinkedBinarySearchTree - Constructors

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 addElement method finds the proper location for the given element and adds it there as a leaf

LinkedBinarySearchTree - addElement

FIGURE 13.3 Adding elements to a binary search tree

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

Removing Elements The removeElement method makes use of a private replacement method to find the proper element to replace a non-leaf element that is removed

LinkedBinarySearchTree - removeElement

LinkedBinarySearchTree - replacement

FIGURE 13.4 Removing elements from a binary tree

Removing All Occurrences The removeAllOccurrences method removes all occurrences of an element from the tree This method uses the removeElement method This method makes a distinction between the first call and successive calls to the removeElement method

LinkedBinarySearchTree - removeAllOccurrences

The removeMin 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

FIGURE 13.5 Removing the minimum element from a binary search tree

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 ListADT and the OrderedListADT in the BinarySearchTreeList class

FIGURE 13.6 The common operations on a list

FIGURE 13.7 The operation particular to an ordered list

FIGURE 13.8 Analysis of linked list and binary search tree implementations of an ordered list

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

FIGURE 13.9 A degenerate binary tree

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))

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

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

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

FIGURE 13.10 Unbalanced tree and balanced tree after a right rotation

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

FIGURE 13.11 Unbalanced tree and balanced tree after a left rotation

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

FIGURE 13.12 A rightleft rotation

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

FIGURE 13.13 A leftright rotation

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

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

FIGURE 13.14 A right rotation in an AVL tree

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

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

FIGURE 13.15 A rightleft rotation in an AVL tree

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 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

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

FIGURE 13.16 Valid red/black trees

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

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

Insertion into Red/Black Trees In the case where the parent of the current node is a right child, there are two cases Leftaunt.color == red Leftaunt.color == black If leftaunt.color is red then the processing steps are:

FIGURE 13.17 Red/black tree after insertion

Insertion into Red/Black Trees If leftaunt.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:

Insertion into Red/Black Trees In the case where the parent of current is a left child, there are two cases: either rightuncle.color == red or rightuncle.color == black If rightuncle.color == red then the processing steps are:

FIGURE 13.18 Red/black tree after insertion

Insertion into Red/Black Trees If rightuncle.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 ot the left around current We then continue as if current was left child to begin with:

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 chid 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

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:

FIGURE 13.19 Red/black tree after removal

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:

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:

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:

Binary Search Trees in the Java Collections API Java provides two implementations of balanced binary search trees TreeSet TreeMap In order to understand the difference between these two, we must first discuss the difference between a set and a map

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

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

The TreeSet and TreeMap Classes Both the TreeSet and TreeMap classes are red/black tree implementations of a balanced binary search tree The operations on both are listed in the following tables

TABLE 13.1 Operations on a TreeSet

TABLE 13.2 Operations on a TreeMap

Ch13 Binary Search Tree

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Ch13 Binary Search Tree