Lab08 - CS61A of UCB(2021-Fall)

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; that is none of the elements is another linked list.

Try to find both an iterative and recursive solution for this problem!

It is easy to solve this problem iteratively. All we have to do is to make a list to store these nodes we have visited while we iterating this linklist.

def convert_link(link):
    """Takes a linked list and returns a Python list with the same elements.
    """
    result = []
    while link is not Link.empty:
        result.append(int(link.first))
        link = link.rest
    return result

The recursive algorithm is also easy. The base case is that when we encounter an empty node, an empty list will be returned. Otherwise, we will break down this problem: current node + the rest of tinkliest

def convert_link(link):
    """Takes a linked list and returns a Python list with the same elements.
    """
    # recursive solution
    if link is Link.empty:
        return []
    else:
        return [int(link.first)] + convert_link(link.rest)

Trees

Q4: Square

Write a function label_squarer that mutates a Tree with numerical labels so that each label is squared.

def label_squarer(t):
    """Mutates a Tree t by squaring all its elements.
    """
    # base case
    if t.is_leaf():
        t.label = t.label ** 2

    # check every branch
    for b in t.branches:
        t.label = t.label ** 2          # change the current node's label
        label_squarer(b)                # change branches

Q5: Cumulative Mul

Write a function cumulative_mul that mutates the Tree t so that each node’s label becomes the product of its label and all labels in the subtrees rooted at the node.

The description indicates that we should make the label of the node equal to the product of its label and all labels in the subtrees in the tree.

The base is the leaf node, which has no subtrees. We will return its label. Otherwise, we check every subtree of the current node. The procedure of changing is bottom-up 🤗

To make the code simple, I use math.prod here. If you also want to use this, you will need to import math at the beginning of the code.

def cumulative_mul(t):
    """Mutates t so that each node's label becomes the product of all labels in
    """
    # base case
    if t.is_leaf():
        return t.label
    # get all label value in subtree
    vals = [cumulative_mul(b) for b in t.branches]

    # calculate
    t.label *= math.prod(vals)

Q6: Add Leaves

Implement add_d_leaves, a function that takes in a Tree instance t and a number v.

We define the depth of a node in t to be the number of edges from the root to that node. The depth of root is therefore 0.

For each node in the tree, you should add d leaves to it, where d is the depth of the node. Every added leaf should have a label of v. If the node at this depth has existing branches, you should add these leaves to the end of that list of branches.

For example, you should be adding 1 leaf with label v to each node at depth 1, 2 leaves to each node at depth 2, and so on.

Here is an example of a tree t(shown on the left) and the result after add_d_leaves is applied with v as 5.

We have to some details into account:

1. How to get the height of the current node? We need to add leaves depending on the height, we can define a helper function with an additional parameter to represent height. Every time we go deeper in the tree, we will +1.
2. How to solve it recursively? The base is the leaf node. We will add leaves to it and return here. Otherwise, we need to repeat the procedure to all its subtrees, and also determine whether the current node needs to add leaves.

def add_d_leaves(t, v):
    """Add d leaves containing v to each node at every depth d.
    """
    def helper(t, v, depth):
        # base case
        if t.is_leaf():
            for i in range(depth):
                t.branches.append(Tree(v))
            return

        # check every branch
        for b in t.branches:
            helper(b, v, depth + 1)
        

        # check current node
        for i in range(depth):
            t.branches.append(Tree(v))

    helper(t, v, 0)

Optional Questions

Q7: 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:

text

>>> 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.

First, If the length of the linklist is less than 2, we will keep it unchanged. Otherwise, we will solve this problem iteratively.

We maintain 2 pointers - last_pos and pos, which indicate the current position and the last position we visited. If the current node is odd-indexed, we let last_pos.rest = post.rest and add the pointers.

def every_other(s):
    """Mutates a linked list so that all the odd-indiced elements are removed
    """
    # if it contains fewer than 2, do nothing
    if s is Link.empty or s.rest is Link.empty:
        return

    last_pos, pos = s, s.rest
    current_index = 1               # start from 2nd position
    while pos is not Link.empty:
        if current_index % 2 == 1: 
            last_pos.rest = pos.rest
        last_pos = pos
        pos = pos.rest
        current_index += 1

Q8: Prune Small

Complete the function prune_small that takes in a Tree t and a number n and prunes t mutatively. If t or any of its branches has more than n branches, the n branches with the smallest labels should be kept and any other branches should be pruned, or removed, from the tree.

Before we start to write the code, we need to think about what is the fast way to prune? The bottom-up way or the up-down way? The latter is the answer. Because we don’t have to check the nodes in the subtrees we have pruned. You may figure out this by observing the template provided in the hint. 🤗

def prune_small(t, n):
    """Prune the tree mutatively, keeping only the n branches
    of each node with the smallest label.
    """
    while len(t.branches) > n:
        largest = max(t.branches, key=lambda x: x.label)
        t.branches.remove(largest)

    for b in t.branches:
        prune_small(b, n)