Lab 8: Linked Lists, Trees
Due by 11:59pm on Tuesday, October 20.
查看汉语翻译
Starter Files
Download lab08.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.
Linked Lists
We've learned that a Python list is one way to store sequential values. Another type of list is a linked list. A Python list stores all of its elements in a single object, and each element can be accessed by using its index. A linked list, on the other hand, is a recursive object that only stores two things: its first value and a reference to the rest of the list, which is another linked list.
We can implement a class, Link
, that represents a linked list object. Each
instance of Link
has two instance attributes, first
and rest
.
class Link:
"""A linked list.
>>> s = Link(1)
>>> s.first
1
>>> s.rest is Link.empty
True
>>> s = Link(2, Link(3, Link(4)))
>>> s.first = 5
>>> s.rest.first = 6
>>> s.rest.rest = Link.empty
>>> s # Displays the contents of repr(s)
Link(5, Link(6))
>>> s.rest = Link(7, Link(Link(8, Link(9))))
>>> s
Link(5, Link(7, Link(Link(8, Link(9)))))
>>> print(s) # Prints str(s)
<5 7 <8 9>>
"""
empty = ()
def __init__(self, first, rest=empty):
assert rest is Link.empty or isinstance(rest, Link)
self.first = first
self.rest = rest
def __repr__(self):
if self.rest is not Link.empty:
rest_repr = ', ' + repr(self.rest)
else:
rest_repr = ''
return 'Link(' + repr(self.first) + rest_repr + ')'
def __str__(self):
string = '<'
while self.rest is not Link.empty:
string += str(self.first) + ' '
self = self.rest
return string + str(self.first) + '>'
A valid linked list can be one of the following:
- An empty linked list (
Link.empty
) - A
Link
object containing the first value of the linked list and a reference to the rest of the linked list
What makes a linked list recursive is that the rest
attribute of a single
Link
instance is another linked list! In the big picture, each Link
instance stores a single value of the list. When multiple Link
s are linked
together through each instance's rest
attribute, an entire sequence is
formed.
Note: This definition means that the
rest
attribute of anyLink
instance must be eitherLink.empty
or anotherLink
instance! This is enforced inLink.__init__
, which raises anAssertionError
if the value passed in forrest
is neither of these things.
To check if a linked list is empty, compare it against the class attribute
Link.empty
. For example, the function below prints out whether or not the
link it is handed is empty:
def test_empty(link):
if link is Link.empty:
print('This linked list is empty!')
else:
print('This linked list is not empty!')
Motivation: Why linked lists
Since you are already familiar with Python's built-in lists, you might be wondering why we are teaching you another list representation. There are historical reasons, along with practical reasons. Later in the course, you'll be programming in Scheme, which is a programming language that uses linked lists for almost everything.
For now, let's compare linked lists and Python lists by looking at two common sequence operations: inserting an item and indexing.
Python's built-in list is like a sequence of containers with indices on them:
Linked lists are a list of items pointing to their neighbors. Notice that there's no explicit index for each item.
Suppose we want to add an item at the head of the list.
- With Python's built-in list, if you want to put an item into the container labeled with index 0, you must move all the items in the list into its neighbor containers to make room for the first item;
- With a linked list, you tell Python that the neighbor of the new item is the old beginning of the list.
Now, let's take a look at indexing. Say we want the item at index 3 from a list.
- In the built-in list, you can use Python list indexing, e.g.
lst[3]
, to easily get the item at index 3. - In the linked list, you need to start at the first item and repeatedly follow
the
rest
attribute, e.g.link.rest.rest.first
. How does this scale if the index you were trying to access was very large?
Can you think of situations where you would want to use one type of list over another? In this class, we aren't too worried about performance. However, in future computer science courses, you'll learn how to make performance tradeoffs in your programs by choosing your data structures carefully.
Trees
Recall that a tree is a recursive abstract data type that has a label
(the
value stored in the root of the tree) and branches
(a list of trees directly
underneath the root).
We saw one way to implement the tree ADT -- using constructor and selector
functions that treat trees as lists. Another, more formal, way to implement the
tree ADT is with a class. Here is part of the class definition for Tree
,
which can be found in lab07.py
:
class Tree:
"""
>>> t = Tree(3, [Tree(2, [Tree(5)]), Tree(4)])
>>> t.label
3
>>> t.branches[0].label
2
>>> t.branches[1].is_leaf()
True
"""
def __init__(self, label, branches=[]):
for b in branches:
assert isinstance(b, Tree)
self.label = label
self.branches = list(branches)
def is_leaf(self):
return not self.branches
Even though this is a new implementation, everything we know about the tree ADT remains true. That means that solving problems involving trees as objects uses the same techniques that we developed when first studying the tree ADT (e.g. we can still use recursion on the branches!). The main difference, aside from syntax, is that tree objects are mutable.
Here is a summary of the differences between the tree ADT implemented using functions and lists vs. implemented using a class:
- | Tree constructor and selector functions | Tree class |
---|---|---|
Constructing a tree | To construct a tree given a label and a list of branches , we call
tree(label, branches)
|
To construct a tree object given a label and a list of branches , we call
Tree(label, branches) (which calls the Tree.__init__ method)
|
Label and branches | To get the label or branches of a tree t , we call label(t) or
branches(t) respectively
|
To get the label or branches of a tree t , we access the instance attributes
t.label or t.branches respectively
|
Mutability | The tree ADT is immutable because we cannot assign values to call expressions | The label and branches attributes of a Tree instance can be
reassigned, mutating the tree |
Checking if a tree is a leaf | To check whether a tree t is a leaf, we call the convenience function
is_leaf(t)
|
To check whether a tree t is a leaf, we call the bound method t.is_leaf() .
This method can only be called on Tree objects. |
Required Questions
What Would Python Display?
Q1: WWPD: Linked Lists
Read over the Link
class in lab08.py
. Make sure you understand the
doctests.
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q link -u
Enter
Function
if you believe the answer is<function ...>
,Error
if it errors, andNothing
if nothing is displayed.If you get stuck, try drawing out the box-and-pointer diagram for the linked list on a piece of paper or loading the
Link
class into the interpreter withpython3 -i lab09.py
.
>>> from lab08 import *
>>> link = Link(1000)
>>> link.first
______1000
>>> link.rest is Link.empty
______True
>>> link = Link(1000, 2000)
______AssertionError
>>> link = Link(1000, Link())
______TypeError
>>> from lab08 import *
>>> link = Link(1, Link(2, Link(3)))
>>> link.first
______1
>>> link.rest.first
______2
>>> link.rest.rest.rest is Link.empty
______True
>>> link.first = 9001
>>> link.first
______9001
>>> link.rest = link.rest.rest
>>> link.rest.first
______3
>>> link = Link(1)
>>> link.rest = link
>>> link.rest.rest.rest.rest.first
______1
>>> link = Link(2, Link(3, Link(4)))
>>> link2 = Link(1, link)
>>> link2.first
______1
>>> link2.rest.first
______2
>>> from lab08 import *
>>> link = Link(5, Link(6, Link(7)))
>>> link # Look at the __repr__ method of Link
______Link(5, Link(6, Link(7)))
>>> print(link) # Look at the __str__ method of Link
______<5 6 7>
Linked Lists
Q2: Convert Link
Write a function convert_link
that takes in a linked list and returns the
sequence as a Python list. You may assume that the input list is shallow; none
of the elements is another linked list.
Try to find both an iterative and recursive solution for this problem!
def convert_link(link):
"""Takes a linked list and returns a Python list with the same elements.
>>> link = Link(1, Link(2, Link(3, Link(4))))
>>> convert_link(link)
[1, 2, 3, 4]
>>> convert_link(Link.empty)
[]
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q convert_link
Q3: Every Other
Implement every_other
, which takes a linked list s
. It mutates s
such
that all of the odd-indexed elements (using 0-based indexing) are removed from
the list. For example:
>>> s = Link('a', Link('b', Link('c', Link('d'))))
>>> every_other(s)
>>> s.first
'a'
>>> s.rest.first
'c'
>>> s.rest.rest is Link.empty
True
If s
contains fewer than two elements, s
remains unchanged.
Do not return anything!
every_other
should mutate the original list.
def every_other(s):
"""Mutates a linked list so that all the odd-indiced elements are removed
(using 0-based indexing).
>>> s = Link(1, Link(2, Link(3, Link(4))))
>>> every_other(s)
>>> s
Link(1, Link(3))
>>> odd_length = Link(5, Link(3, Link(1)))
>>> every_other(odd_length)
>>> odd_length
Link(5, Link(1))
>>> singleton = Link(4)
>>> every_other(singleton)
>>> singleton
Link(4)
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q every_other
Trees
Q4: Cumulative Mul
Write a function cumulative_mul
that mutates the Tree t
so that each node's
label becomes the product of all labels in the subtree rooted at the node.
def cumulative_mul(t):
"""Mutates t so that each node's label becomes the product of all labels in
the corresponding subtree rooted at t.
>>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
>>> cumulative_mul(t)
>>> t
Tree(105, [Tree(15, [Tree(5)]), Tree(7)])
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q cumulative_mul
Optional Problems
Q5: Cycles
The Link
class can represent lists with cycles. That is, a list may
contain itself as a sublist.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> s.rest.rest.rest.rest.rest.first
3
Implement has_cycle
,that returns whether its argument, a Link
instance, contains a cycle.
Hint: Iterate through the linked list and try keeping track of which
Link
objects you've already seen.
def has_cycle(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle(t)
False
>>> u = Link(2, Link(2, Link(2)))
>>> has_cycle(u)
False
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q has_cycle
As an extra challenge, implement has_cycle_constant
with only constant space. (If you
followed
the hint above, you will use linear space.) The solution is short (less than 20
lines of code), but requires a clever idea. Try to discover the solution
yourself before asking around:
def has_cycle_constant(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle_constant(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle_constant(t)
False
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q has_cycle_constant
Q6: Reverse Other
Write a function reverse_other
that mutates the tree such that labels on
every other (odd-depth) level are reversed. For example,
Tree(1,[Tree(2, [Tree(4)]), Tree(3)])
becomes Tree(1,[Tree(3, [Tree(4)]), Tree(2)])
.
Notice that the nodes themselves are not reversed; only the labels are.
def reverse_other(t):
"""Mutates the tree such that nodes on every other (odd-depth) level
have the labels of their branches all reversed.
>>> t = Tree(1, [Tree(2), Tree(3), Tree(4)])
>>> reverse_other(t)
>>> t
Tree(1, [Tree(4), Tree(3), Tree(2)])
>>> t = Tree(1, [Tree(2, [Tree(3, [Tree(4), Tree(5)]), Tree(6, [Tree(7)])]), Tree(8)])
>>> reverse_other(t)
>>> t
Tree(1, [Tree(8, [Tree(3, [Tree(5), Tree(4)]), Tree(6, [Tree(7)])]), Tree(2)])
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q reverse_other