# Lab: Binary search trees

Summary
We explore binary search trees and their use in implementing the Map ADT.
Repository
https://github.com/Grinnell-CSC207/lab-bsts-2019

## Preparation

a. Fork and clone the specified respository.

b. Import the repository into Eclipse.

c. Skim the code to make sure that you understand the structure of the BST data structure.

## Exercises

### Exercise 1: Getting elements

As you likely noted, a BSTNode has four fields: a key, a value, a pointer to the left subtree, and a pointer to the right subtree.

We compare values in the tree using the comparator field in the SimpleBST class.

a. Review the documentation for get(K key) in SimpleMap.java.

b. Sketch how you would write the get(K key) method for the SimpleBST.java class.

c. Compare your answer to the extant get method.

### Exercise 2: Our experiment

a. Review BSTExperiment.java and note to yourself what the experiment checks and what output you would expect.

b. Run the code to see if you get the expected output.

### Exercise 3: Setting elements

Unfortunately, we have not yet implemented the set(K key, V value) method. There are two typical approaches for set, one iterative and one recursive.

In the iterative approach, you repeatedly follow the appropriate branch to the left or right subtree until either (a) you find a node that contains the same key or (b) you observe that the subtree you are about to enter is null. In case (a), you replace the corresponding value and do not change the size of the tree. In case (b), you build a new node, set the appropriate link, and increment the size. You will also need a special case for the root.

In the recursive approach, you write a helper procedure that takes a node in addition to the key and value as parameters and returns a modified subtree. If the node is null, you build a new node. If the key is smaller than the node’s key, you recursively set the left subtree to the result of calling the helper procedure on the left subtree. If the key is larger than the node’s key, you recursively set the right subtree to the result of calling the helper procedure on the right subtree. (This approach also works well if you decide to have immutable trees.) You don’t need a special case for the root, since you can just write root = helper(root, key, value). However, you may need another field to keep track of the old value associated with the key, if there is one. (Fortunately, our SimpleBST class has a cachedValue field for just that purpose.)

a. Review the documentation for the set method in SimpleMap.java.

b. You should have received a card from your instructor noting that you are in group I(terative) or group R(ecursive). Implement the set method using the specified approach. Be prepared to discuss issues with your peers.

c. Rerun the experiment to determine whether your set method seems to be operating correctly.

### Exercise 4: Applying a procedure to each entry

a. Review the documentation for forEach.

b. Implement forEach.

### Excercise 5: Iteration

Finish implementing nodes().

## For those with extra time

If you find that you have extra time, try any of the following problems.

### Extra 1: An alternate set

a. You implemented set iteratively or recursively. Implement it in the other way.

b. Which implementation do you prefer? Why?

Revise forEach so that it iterates values in a breadth-first rather than depth-first order. (If you already had it iterate in a breadth-first order, have it iterate in a depth-first order.)

### Extra 3: Removing elements

Sketch how you might implement the primary remove method.

## Notes

If you need to simulate a recursive method without recursion, the most common strategy is to use a stack.

The typical way to do breadth-first iteration is with a queue.

If you are assigned the iterative version of set, here’s pseudocode

if (root is null)
set the root to a new node
else
set current to root
while (...)
compare key to current.key
case =:
replace the value
return the old value
case <:
if there is a left subtree,
current = left
otherwise
current.left = new node
return null
case >:
if there is a right subtree,
current = right
otherwise
current.right = new node
return null


If you are assigned the recursive version of set, here’s pseudocode.

function set(key, value)
root = setHelper(root, key, value)
return cache

function setHelper(node, key, value)
if node == null
set cache to null
return a new node
else
compare key to node.key
case =:
set cache to value in node
update value in node
return node
case <:
node.left = setHelper(node.left, key, value);
return node;
else
node.right = setHelper(node.right, key, value);
return node;