CS275

Due Friday, October 7

For this assignment you will write a number of programs that involve trees. Recall that a tree is a data structure that is either empty, or contains some data and has 0 or more children, each of which is a tree.

We will use Scheme procedures to implement the data type.At the beginning of the following section the tree datatype will be defined. All of this code is contained in the file "TreeDatatype.rkt". You should put your solutions to the exercises in a new file which starts

#lang racket

(require "TreeDatatype.rkt")

The goals for this assignment are to

- Give you a deeper understanding of functions
- Give you a deeper understanding of recursion
- Give you a deeper understanding of datatypes and how to work with them in Scheme

You will need to download the file TreeDatatype.rkt

A tree is either

- an empty tree, which has no fields
- a non-empty tree, which has a
*value*field containing a number and a*list-of-children*field containing a list of children of the node, which are themselves trees.

(define tree? (lambda (t) (or (empty-tree? t) (non-empty-tree? t))))

(define empty-tree? (lambda (x) (null? x)))

(define non-empty-tree? (lambda (x)

(cond

[(list? x) (eq? (car x) 'tree)]

[else #f])))

; constructors

(define non-empty-tree (lambda (value list-of-children)

(list 'tree value list-of-children)))

(define empty-tree (lambda () null))

; Convenience constructor

(define makeTree (lambda l

(non-empty-tree (car l) (cdr l))))

; utility functions

(define value (lambda (tree)

(cond

[(empty-tree? tree) 'error]

[else (cadr tree)])))

(define list-of-children (lambda (tree)

(cond

[(empty-tree? tree) 'error]

[else (caddr tree)])))

(define number-of-children (lambda (tr)

(cond

[(empty-tree? tree) 'error]

[else (length (list-of-children tr))])))

(define leaf? (lambda (tr)

(cond

[(empty-tree? tree) 'error]

[else (zero? (number-of-children tr))])))

; example trees

(define Empty (empty-tree))

(define T1 (makeTree 50))

(define T2 (makeTree 22))

(define T3 (makeTree 10))

(define T4 (makeTree 5))

(define T5 (makeTree 17))

(define T6 (makeTree 73 T1 T2 T3))

(define T7 (makeTree 100 T4 T5))

(define T8 (makeTree 16 T6 T7))

For example, here is a picture of T8 showing the others as subtrees:

We can view T8 as a list: (16 (73 (50) (22) (10)) (100 (5) (17))). Be sure that you can see how the tree structure is preserved in this list.

All of this code is in file TreeDatatype.rkt

In the exercises that follow, you should use higher-order functions like **map,** **apply**, and **fold** to make your solution as concise as possible. You should also use **letrec**to define any recursive help functions.

All programs should be written using the tree operators defined above. Don't directly manipulate the list representations; use the tree procedures.

As usual, if the solution to an earlier problem is useful in a later one, go ahead and use it.

In all of these tr is a value of the tree datatype

## Exercise 1: (childSum tr)

This returns the sum of the values in hte children of the given node.. The tree is assumed to be non-empty.

(childSum T8) returns 173 (childSum T6) returns 82

## Exercise 2: (allSum tr)

This sums the entire set of values found in the tree. If tree is empty it returns 0.

(allSum T8) returns 293 (allSum T6) returns 155

## Exercise 3: (tree->list tr )

This converts a tree into a list format

(tree->list T8) returns (16 (73 (50) (22) (10)) (100 (5) (17)))

## Exercise 4: (list->tree l)

This is the inverse of tree->list.

(tree->list (list-> tree '(16 (73 (50) (22) (10)) (100 (5) (17))))) returns (16 (73 (50) (22) (10)) (100 (5) (17)))

## Exercise 5: (visitTree f tr)

Here f is a function of 1 numerical variable. This returns a new tree with the same structure only every value

vin tree has been transformed to(f v).

(tree->list (visitTree (lambda (x) (* x 3)) T1)) returns (150) (tree->list (visitTree (lambda (x) (* x 3)) T8)) returns (48 (219 (150) (66) (30)) (300 (15) (51)))

## Exercise 6: (sizeof tr)

This returns the number of elements in the subtree rooted at the given node.

(sizeof T6)) returns 4 (sizeof T8) returns 8

## Exercise 7: (height tr)

Returns the height of the tree. Assume that the empty tree has height 0, a single-node tree has height 1, etc.

(height T1) returns 1 (height T8) returns 3

## Exercise 8: (count n tr)

Returns the number of times value n appears in the tree.

(count 22 T8) returns 1 (count 7 T8) returns 0

## Exercise 9: (preorder tr) and Exercise 10: (postorder tr)

Each of these produces a flat list of the values in the tree. They differ only in the order in which the values are listed:

- (preorder tr) lists the value at the root ahead of the values of the children.

(preorder T8) returns (16 73 50 22 10 100 5 17)- (postorder tr) lists the value at the root after the values of the children.

(postorder T8) returns (50 22 10 73 5 17 100 16)