## Introduction

As usual, pull the files from the skeleton and make a new IntelliJ project.

## Exercise: Collections

An important operation provided by `Collection`

classes is the `contains`

method: finding whether a given item is in the `Collection`

. For certain tasks in which this operation needs to be efficient, we need to choose a good data structure.

Let’s take a look at some of the data structures we’ve already looked at. For each of the questions below, the answer is one of the five options:

- a constant number
- proportional to
- proportional to
- proportional to
- proportional to

Submit your answers to the Gradescope submission.

### Exercise 1

Suppose that integers are stored in an `IntList`

object. How many comparisons are necessary in the worst case to determine if a given integer occurs in the sequence?

### Exercise 2

Suppose that comparable objects are stored in an `ArrayList`

. How many comparisons are necessary in the worst case to determine if a given object occurs in the list?

### Exercise 3

Suppose that integers are stored in *increasing* order in an `IntList`

object. How many comparisons are necessary in the worst case to determine if a given object occurs in the list?

### Exercise 4

Suppose that integers are stored in *increasing* order in an array. How many comparisons are necessary in the worst case to determine if a given integer occurs in the sequence?

## Binary Search

In the last question above, we achieve an improved runtime by employing the well known divide-and-conquer algorithm known as **binary search**. Used with an array where `low`

, `mid`

, and `high`

are array indices, it assumes that the elements of the array are in increasing order, and executes the following:

- Set
`low`

to 0 and`high`

to the length of the array minus 1. The value we’re looking for — we’ll call it`k`

— will be somewhere between position`low`

and position`high`

if it’s in the array. - While
`low`

`high`

, do one of the following:- Compute
`mid`

, the*middle*of the range`[low, high]`

, and see if that’s`k`

. If so, return`mid`

. - Otherwise, we can cut the range of possible positions for
`k`

in half, by setting`high`

to`mid - 1`

or by setting`low`

to`mid + 1`

, depending on the result of the comparison.

- Compute
- If the loop terminates with
`low > high`

, we know that`k`

is not in the array, so we return some indication of failure.

The diagrams below portray a search if `k`

was equal to 25. Elements removed from consideration at each iteration are greyed out.

`low = 0`

,`high = 14`

,`mid = 7`

`low = 0`

,`high = 6`

,`mid = 3`

`low = 4`

,`high = 6`

,`mid = 5`

`low = 4`

,`high = 4`

,`mid = 4`

Since (roughly) half the elements are removed from consideration at each step, the worst-case running time is proportional to , where is the number of elements in the array.

## Exercise: Binary Search

Consider applying the binary search algorithm to a sorted doubly linked list (a list where each node has references to both its previous and next node). The variables `low`

, `high`

, and `mid`

would then be references to nodes in the list. Will this work?

In fact, binary search will not work in any linked list structure because we do not have a quick way of moving the `low`

, `high`

, or `mid`

references. In addition, the `contains`

operation would have to take time proportional to the number of elements in the collection. Which of the steps in the binary search algorithm would be the bottleneck that causes a decrease in performance in a sorted doubly linked list, relative to a sorted array for the `contains`

operation?

- Step 1, because it takes extra time to figure out the length of the linked-list
- Step 2a, because it takes extra time to check if we have found the key
- Step 2a, because it takes extra time to calculate the index of the middle node
- Step 2a, because takes extra time to access the middle nodes

Discuss with your partner, and mark your answer in the Gradescope assessment.

## Binary Search Trees

The binary search algorithm suggests a way to organize keys in an explicitly linked tree, as indicated in the diagram below.

The data structure that results is called a **binary search tree** (BST). Given that the root value (one of the keys to be stored) is , a binary search tree is organized as follows:

- Put all the keys that are smaller than into a binary search tree, and let that tree be ’s left subtree.
- Put all the keys that are larger than into a binary search tree, and let that tree be ’s right subtree.

This organization assumes that there are no duplicate keys among those to be stored.

USFCA put together a BST visualization interactive animation to help you visualize the BST insertion and deletion algorithms. Try inserting a key that ends up as a right child and another key that ends up as a left child. Add some more nodes to the tree, then delete some of them: a node with no children, a node with one child, and a node with two children.

Note that this animation removes from the BST by swapping with the inorder *predecessor* rather than the inorder successor. Convince yourself this is essentially equivalent.

## Exercise: BST Identification

Now, let’s practice identifying whether certain binary trees are binary search trees. For each of the following questions, discuss with your partner what you think the solution is. Then, put your answers in the Gradescope assessment as either:

- space separated lists of numbers
- “None”

### Exercise 1

What nodes (can be multiple) must be removed from this tree to make it a binary search tree?

### Exercise 2

What nodes (can be multiple) must be removed from this tree to make it a binary search tree?

### Exercise 3

What nodes (can be multiple) must be removed from this tree to make it a binary search tree?

`contains`

It’s important to note that in a binary search tree, each subtree is also a binary search tree. This suggests a recursive or iterative approach for implementing many methods, and the `contains`

method is no different. In pseudocode, here is an outline of the helper method of the `contains`

method, `containsHelper(TreeNode t, T key)`

:

- An empty tree cannot contain anything, so if
`t`

is`null`

return`false`

. - If
`key`

is equal to`t.item`

, return`true`

. - If
`key < t.item`

,`key`

must be in the left subtree if it’s in the structure at all, so return the result of searching for it in the left subtree. - Otherwise it must be in the right subtree, so return the result of searching for
`key`

in the right subtree.

Note: that the type of `key`

is `T`

, which is the generic type of the `BinaryTree`

class.

This algorithm can go all the way down to a leaf to determine its answer. Thus in the worst case, the number of comparisons is proportional to , the depth of the tree. In a balanced tree (more on that next lab), you can expect the depth of the tree to be proportional to in the worst case, where is the number of nodes in the tree.

### Use of `Comparable`

objects

Finding a value in the tree will require “less than”, “greater than”, and “equals” comparisons. Since the operators < and > don’t work with objects, we have to use method calls for comparisons.

The Java convention for this situation is to have the values stored in the tree be objects that implement the `Comparable`

interface. This interface prescribes just one method, `int compareTo (Object)`

. For `Comparable`

objects `obj1`

and `obj2`

, `obj1.compareTo (obj2)`

returns

- negative if
`obj1`

is less than`obj2`

, - positive if
`obj1`

is greater than`obj2`

, and - zero if the two objects have equal values.

### Balance and Imbalance

Unfortunately, use of a binary search tree does not guarantee efficient search. For example, the tree

is a binary search tree in which search proceeds in the same runtime as a linked list. We thus are forced to consider the *balance* of a binary search tree. Informally, a balanced tree has subtrees that are roughly equal in size and depth. Next lab, we will encounter specific algorithms for maintaining balance in a binary search tree. Until then, we will work under the possibly unwarranted assumption that we don’t need to worry much about balance.

One can prove, incidentally, that search in a BST of keys will require only about comparisons (where is the “natural log” of ) if the keys are inserted in random order. Well-balanced trees are common, and degenerate trees are rare.

### Insertion into a BST

If we have 4 nodes in our binary search tree, there are actually 14 different BST’s you could make. Correspondingly, there are typically a bunch of places in a BST that a key to be inserted might go, anywhere from the root down to a leaf. Given below are two trees; the tree on the right shows one possible way to insert the key 41 into the tree on the left.

However, to minimize restructuring of the tree and the creation of internal nodes, we choose in the following exercise to insert a new key only as a new *leaf*.

## Exercise: BST Implementation

Now it’s time to start writing code! As you go, don’t forget to write JUnit tests.

Since binary search trees share many of the characteristics of regular binary trees, we can define the `BinarySearchTree`

class using inheritance from a provided `BinaryTree`

class.

### Exercise 1: Implement Constructors

The `BinarySearchTree`

class is defined as follows:

```
public class BinarySearchTree<T extends Comparable<T>> extends BinaryTree<T>
```

This class definition is a slightly more complicated use of generic types than you have seen before in lab. Previously, you saw things like `BinaryTree<T>`

, which meant that the `BinaryTree`

class had a generic type `T`

that could be any class, interface, or abstract class that extends `Object`

. In this case, `BinarySearchTree<T extends Comparable<T>>`

means that the `BinarySearchTree`

class has a generic type `T`

that can be any class, interface, or abstract class that implements the `Comparable<T>`

interface. In this case, `Comparable<T>`

is used because the `Comparable`

interface itself uses generic types (much like the `Iterable`

and `Iterator`

interfaces).

Take a look through the `BinarySearchTree`

and `BinaryTree`

classes, and familiarize yourself with the methods that are available to you. Then, implement the two constructors for the `BinarySearchTree`

class, using only method calls to `super`

.

### Exercise 2: `contains`

Now, we will implement the `contains`

method. We will use the following method signature:

```
public boolean contains(T key)
```

which takes a `Comparable`

object `T`

as an argument and checks whether the tree contains it.

Recall that `Comparable`

objects provide an `int compareTo`

method that returns:

- a negative integer if this object is less than the argument,
- a positive integer if this object is greater than the argument, and
- 0 if the two objects have equal values.

Depending on whether you take a recursive or iterative approach, you may need to define a helper method. If you’re stuck, take a look at the pseudocode that we described above!

### Exercise 3: `add`

We will now define an `add`

method. We will use the following method signature:

```
public void add(T key)
```

which takes a `Comparable`

object as an argument and adds it to the tree *if and only if it isn’t already there*. The trees you create with the `add`

method will thus not contain any duplicate elements.

*Hint*: You should be able to do this in a similar way to the `contains`

method. When you’re done with both, you can write a JUnit test suite to test your code. Don’t forget edge cases!

### Exercise 4: Comparison Counting

How many comparisons *between keys* are necessary to produce the sample tree shown below? Ignore equality checks. Put your answer in the Gradescope assessment.

Remember: “However, to minimize restructuring of the tree and the creation of internal nodes, we choose in the following exercises to insert a new key only as a new leaf.”

### Optional Exercise 5: Pre-in-order

Our next (optional) function will be in the `BinaryTree`

class rather than the `BinarySearchTree`

class. Note that past CS 61BL students have found this exercise quite difficult, but **we still strongly recommended that you complete this exercise to expand your understanding of binary tree structures**.

Write and test a constructor for the `BinaryTree`

class that, given two lists, constructs the corresponding binary tree (**not necessarily a BST**). The first list contains the objects in the order they would be enumerated in a *preorder* traversal; the second list contains the objects in the order they would be enumerated in an *inorder* traversal.

For example, given `["A", "B", "C", "D", "E", "F"]`

(preorder) and `["B", "A", "E", "D", "F", "C"]`

(inorder), you should initialize the tree as follows:

You may assume every item in the tree is unique when doing this exercise.

## Discussion: BST Deletion

We’ve covered `add`

ing to a BST and `contains`

in a BST. But how about deletion?

When inserting a key, we were allowed to choose where in the tree it should go. The deletion operation doesn’t appear to allow us that flexibility; deletion of some keys will be easy (leaves or keys with only one child), but our deletion method has to be able to delete *any* key from the tree.

Here are a bunch of binary search trees that might result from deleting the root of the tree

Which ones do you think are reasonable?

### A Good Way to Delete a Key

The following algorithm for deletion has the advantage of minimizing restructuring and unbalancing of the tree. The method returns the tree that results from the removal.

- Find the node to be removed. We’ll call it
`remNode`

. If it’s a leaf, remove it immediately by returning`null`

. If it has only one child, remove it by returning the other child node. - Otherwise, remove the
*inorder successor*of`remNode`

, copy its`item`

into`remNode`

, and return`remNode`

.

What is an *inorder successor*? It is the node that would appear AFTER the `remNode`

if you were to do an inorder traversal of the tree.

An example is diagrammed below. The node to remove is the node containing 4. It has two children, so it’s not an easy node to remove. We locate the node with 4’s inorder successor, namely 5. The node containing 5 has no children, so it’s easy to delete. We copy the 5 into the node that formerly contained 4, and delete the node that originally contained 5.

Before | After |
---|---|

### Inorder Successor

Suppose `node`

is the root node in a BST with both a left child and a right child. Will `sucNode`

, the inorder successor of `node`

, ALWAYS have a null left child? Discuss this with your partner.

We’ve implemented a `delete`

method for you already. Take a look at it and **understand how it works**.

## The Engineer’s Tradeoff

Consider the problem of finding the `k`

th largest key in a binary search tree. An obvious way of doing this is to use the inorder enumeration from this week’s lab; we could call `nextElement`

until we get the desired key. If `k`

is near the number of keys `N`

in the tree, however, that algorithm’s running time is proportional to `N`

. We’d like to do better. But how?

If you haven’t yet noticed, this class is all about tradeoffs—finding the delicate balance between two conflicting factors to perfectly suit a certain task. Choosing an appropriate data structure is one tradeoff: an algorithm that requires quick access to a certain piece of information would perform better on an array, but an algorithm that uses many insert and delete operations would probably do better in a linked list. Sacrificing a shorter running time in exchange for more memory space, or vice versa, is another.

For this problem, we can reduce the runtime of the `k`

th largest key by storing in each node the size of the tree rooted at that node. Can you design an algorithm using this idea that runs in time proportional to `d`

, where `d`

is the depth of the tree?

## Conclusion

### Submission

- Complete the following methods in
`BinarySearchTree.java`

:`BinarySearchTree()`

`BinarySearchTree(TreeNode root)`

`contains(T key)`

`add(T key)`

- Submit the Gradescope assessment.
- Understand the
`delete`

method of the`BinarySearchTree`

. - Optionally (but highly recommended), complete the constructor in
`BinaryTree.java`

. - Complete the self-reflection linked in the Gradescope assessment.