- Assigned
- Friday, 19 April 2024
- Summary
- We explore binary search trees and their use in implementing the Map ADT.

a. Fork and clone the respository at https://github.com/Grinnell-CSC207/lab-bsts.

b. Import the repository into VSCode.

c. Skim the `BSTNode`

class and the fields and constructors in
`SimpleBST`

to make sure that you understand the organiation of the
BST data structure.

*Driver: A*

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 in `SimpleBST.java`

.

*Driver: B*

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.

*Driver: B*

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.

*Driver: A*

a. Review the documentation for `forEach`

.

b. Implement `forEach`

.

*Driver: B*

Finish implementing `nodes()`

.

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

`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.)

Sketch how you might implement the primary `remove`

method.

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
add 1 to size
return null
case > (the "else case"):
if there is a right subtree,
current = right
otherwise
current.right = new node
add 1 to size
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
add 1 to size
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;
case >: (the "else case")
node.right = setHelper(node.right, key, value);
return node;
```