# CS 61A

Time: Thu 10/12/17 3 pm

## Misc about Trees

They are naturally sustainable! Oh well, think about how trees are conceptually related to pairs. Define the terminologies around trees--tree, root, label, branch, leaf (the language we use may vary in different places).

Consider the following implementation:

``````def tree(label, branches=[]):
return [label] + branches

def label(tree):
return tree[0]

def branches(tree):
return tree[1:]``````

There are multiple ways to get started on playing with such tree structure! As a few examples:

• Track 1
1. Given a tree, how can we find a largest/smallest label in the leaves?
2. Given a tree, how can we find a largest/smallest label in the tree?
3. Given a tree, how can we find the sum of all labels (assuming all labels are numbers)?
• Track 2
1. Given a tree, how can we clone it?
2. Given a tree, how can we make a copy of it with all the leaves pruned?
• Track 3
1. Given a tree and a label, how can we check if the label we want is in the tree?
2. Given a tree, how can we check if there aren't any descendent nodes of a same label for any node in the tree?

## Midterm 2 Practice

### This One Goes to Eleven (fa14-mt2-3)

1. Fill in the blanks of the implementation of `sixty_ones` below, a function that takes a `Link` instance representing a sequence of integers and returns the number of times that 6 and 1 appear consecutively.

``````def sixty_ones(s):
"""
Return the number of times that 1 directly follows 6 in linked list s.

>>> once = Link(4, Link(6, Link(1, Link(6, Link(0, Link(1))))))
>>> twice = Link(1, Link(6, Link(1, once)))
>>> thrice = Link(6, twice)
>>> apply_to_all(sixty_ones, [Link.empty, once, twice, thrice])
[0, 1, 2, 3]
"""

if :
return 0
elif :
return 1 +
else:
return ``````
2. Fill in the blanks of the implementation of `no_eleven` below, a function that returns a list of all distinct length-n lists of ones and sixes in which 1 and 1 do not appear consecutively.

``````def no_eleven(n):
"""
Return a list of lists of 1’s and 6’s that do not contain 1 after 1.

>>> no_eleven(2)
[[6, 6], [6, 1], [1, 6]]
>>> no_eleven(3)
[[6, 6, 6], [6, 6, 1], [6, 1, 6], [1, 6, 6], [1, 6, 1]]
>>> no_eleven(4)[:4] # first half
[[6, 6, 6, 6], [6, 6, 6, 1], [6, 6, 1, 6], [6, 1, 6, 6]]
>>> no_eleven(4)[4:] # second half
[[6, 1, 6, 1], [1, 6, 6, 6], [1, 6, 6, 1], [1, 6, 1, 6]]
"""

if n == 0:
return
elif n == 1:
return
else:
a, b = no_eleven(), no_eleven()
return [ for s in a] + [ for s in b]``````

### Game of Thrones (su16-mt-7)

This question uses the following tree data abstraction.

``````def tree(entry, children=[]):
return [entry, children]

def entry(tree):
return tree[0]

def children(tree):
return tree[1]``````
1. Define the function `track_lineage` that takes in a tree of strings `family_tree` and a string `name`. Assume that there is a unique node with entry `name`. `track_lineage` returns a list with the entries of the parent and grandparent of that node. If the node with entry name does not have a parent or grandparent, return `None` for that element in the list. See the doctests for details. Do not violate abstraction barriers. You may only use the lines provided. You may not need to fill all the lines.

``````def track_lineage(family_tree, name):
"""
Return the entries of the parent and grandparent of the node with entry name in family_tree.

>>> t = tree('Tytos', [
...   tree('Tywin', [
...     tree('Cersei'), tree('Jaime'), tree('Tyrion')
...   ]),
...   tree('Kevan', [
...     tree('Lancel'), tree('Martyn'), tree('Willem')
...   ])])
>>> track_lineage(t, 'Tywin')
['Tytos', None]
>>> track_lineage(t, 'Tytos')
[None, None]
"""

def tracker(t, p, gp):
if

for c in children(t):

return tracker(, , )``````
2. Assuming that `track_lineage` works correctly, define the function are cousins that takes in a tree of strings `family_tree` and two strings `name1` and `name2` and returns `True` if the node with entry `name1` and the node with entry `name2` are cousins in `family_tree`. Assume that there are unique nodes with entries `name1` and `name2` in `family_tree`. See the doctests for details. Two nodes are cousins if they have the same grandparent but different parents.

You may only use the lines provided. You may not need to fill all the lines.

``````def are_cousins(family_tree, name1, name2):
"""Return True if a node with entry name1 is a cousin of a node with entry name2 in family_tree.
>>> are_cousins(t, ‘Kevan’, ‘Tytos’) # same tree as before False
>>> are_cousins(t, ‘Cersei’, ‘Lancel’)
True
>>> are_cousins(t, ‘Jaime’, ‘Lancel’)
True
>>> are_cousins(t, ‘Jaime’, ‘Tyrion’)
False
"""

``````

### Tree Time (fa14-mt2-4)

1. A `GrootTree` `g` is a binary tree that has an attribute `parent`. Its parent is the `GrootTree` in which `g` is a branch. If a `GrootTree` instance is not a branch of any other `GrootTree` instance, then its parent is `BinaryTree.empty`.

`BinaryTree.empty` should not have a `parent` attribute. Assume that every `GrootTree` instance is a branch of at most one other `GrootTree` instance and not a branch of any other kind of tree.

Fill in the blanks below so that the `parent` attribute is set correctly. You may not need to use all of the lines. Indentation is allowed. You should not include any `assert` statements. Using your solution, the doctests for `fib_groot` should pass.

``````class GrootTree(BinaryTree):
"""A binary tree with a parent."""

def __init__(self, entry, left=BinaryTree.empty, right=BinaryTree.empty):
BinaryTree.__init__(self, entry, left, right)

def fib_groot(n):
"""
Return a Fibonacci GrootTree.

>>> t = fib_groot(3)
>>> t.entry
2
>>> t.parent.is_empty
True
>>> t.left.parent.entry
2
>>> t.right.left.parent.right.parent.entry
1
"""

if n == 0 or n == 1:
return GrootTree(n)
else:
left, right = fib_groot(n - 2), fib_groot(n - 1)
return GrootTree(left.entry + right.entry, left, right)``````
2. Fill in the blanks of the implementation of `paths`, a function that takes two arguments: a `GrootTree` instance `g` and a list `s`. It returns the number of paths through `g` whose entries are the elements of `s`. A path through a `GrootTree` can extend either to a branch or its `parent`.

You may assume that the `GrootTree` class is implemented correctly and that the list `s` is non-empty.

There are two paths for `[2, 1, 2, 1, 0]`, one path for `[2, 1, 0, 1, 0]` in the tree below:

```    2
+-------+
1       1
+---+
0   1```

``````def paths(g, s):
"""
The number of paths through g with entries s.

>>> t = fib_groot(3)
>>> paths(t, [1])
0
>>> paths(t, [2])
1
>>> paths(t, [2, 1, 2, 1, 0])
2
>>> paths(t, [2, 1, 0, 1, 0])
1
>>> paths(t, [2, 1, 2, 1, 2, 1])
8
"""

if g is BinaryTree.empty :
return 0
elif :
return 1
else:
xs = []
return sum([ for x in xs])``````