Python inheriting base class variable [duplicate] - python

I'm trying to simplify one of my homework problems and make the code a little better. What I'm working with is a binary search tree. Right now I have a function in my Tree() class that finds all the elements and puts them into a list.
tree = Tree()
#insert a bunch of items into tree
then I use my makeList() function to take all the nodes from the tree and puts them in a list.
To call the makeList() function, I do tree.makeList(tree.root). To me this seems a little repetitive. I'm already calling the tree object with tree.so the tree.root is just a waste of a little typing.
Right now the makeList function is:
def makeList(self, aNode):
if aNode is None:
return []
return [aNode.data] + self.makeList(aNode.lChild) + self.makeList(aNode.rChild)
I would like to make the aNode input a default parameter such as aNode = self.root (which does not work) that way I could run the function with this, tree.makeList().
First question is, why doesn't that work?
Second question is, is there a way that it can work? As you can see the makeList() function is recursive so I cannot define anything at the beginning of the function or I get an infinite loop.
EDIT
Here is all the code as requested:
class Node(object):
def __init__(self, data):
self.data = data
self.lChild = None
self.rChild = None
class Tree(object):
def __init__(self):
self.root = None
def __str__(self):
current = self.root
def isEmpty(self):
if self.root == None:
return True
else:
return False
def insert (self, item):
newNode = Node (item)
current = self.root
parent = self.root
if self.root == None:
self.root = newNode
else:
while current != None:
parent = current
if item < current.data:
current = current.lChild
else:
current = current.rChild
if item < parent.data:
parent.lChild = newNode
else:
parent.rChild = newNode
def inOrder(self, aNode):
if aNode != None:
self.inOrder(aNode.lChild)
print aNode.data
self.inOrder(aNode.rChild)
def makeList(self, aNode):
if aNode is None:
return []
return [aNode.data] + self.makeList(aNode.lChild) + self.makeList(aNode.rChild)
def isSimilar(self, n, m):
nList = self.makeList(n.root)
mList = self.makeList(m.root)
print mList == nList

larsmans answered your first question
For your second question, can you simply look before you leap to avoid recursion?
def makeList(self, aNode=None):
if aNode is None:
aNode = self.root
treeaslist = [aNode.data]
if aNode.lChild:
treeaslist.extend(self.makeList(aNode.lChild))
if aNode.rChild:
treeaslist.extend(self.makeList(aNode.rChild))
return treeaslist

It doesn't work because default arguments are evaluated at function definition time, not at call time:
def f(lst = []):
lst.append(1)
return lst
print(f()) # prints [1]
print(f()) # prints [1, 1]
The common strategy is to use a None default parameter. If None is a valid value, use a singleton sentinel:
NOTHING = object()
def f(arg = NOTHING):
if arg is NOTHING:
# no argument
# etc.

If you want to treat None as a valid argument, you could use a **kwarg parameter.
def function(arg1, arg2, **kwargs):
kwargs.setdefault('arg3', default)
arg3 = kwargs['arg3']
# Continue with function
function("amazing", "fantastic") # uses default
function("foo", "bar", arg3=None) # Not default, but None
function("hello", "world", arg3="!!!")
I have also seen ... or some other singleton be used like this.
def function(arg1, arg2=...):
if arg2 is ...:
arg2 = default

Related

Delete min binary search tree Python

I am trying to delete the minimum node from a BST, so I search through the tree until I get the min (when root.leftnode is None) and then set root.rightnode to the root itself to continue the BST.
The issue is when I check the tree after doing this it does not show the deletion ever occurred.
Could someone point me in the right direction please, any advice is appreciated.
class node():
def __init__(self, key, data):
self.data = data
self.key = key
self.leftnode = None
self.rightnode = None
self.count = 1
class binarysearch():
def __init__(self):
self.size = 0
self.rootnode = None
def insert(self, key, data):
if self.rootnode is None:
self.rootnode = node(key, data)
else:
self.insertnode(self.rootnode, key, data)
def getroot(self):
return self.rootnode
def insertnode(self, root, key, data):
if root.key == key:
root.data = data
elif key < root.key:
if root.leftnode is None:
root.leftnode = node(key, data)
else:
self.insertnode(root.leftnode, key, data)
else:
if root.rightnode is None:
root.rightnode = node(key, data)
else:
self.insertnode(root.rightnode, key, data)
root.count = 1 + self.sizenode(root.leftnode) + self.sizenode(root.rightnode)
def inorder(self, root):
if root is not None:
self.inorder(root.leftnode)
print(root.key)
self.inorder(root.rightnode)
def deletemin(self):
if self.rootnode is None:
print("No nodes exist")
else:
self.deleteminnode(self.rootnode.leftnode)
def deleteminnode(self, root):
if root.leftnode is not None:
self.deleteminnode(root.leftnode)
else:
print (root.key, "deleted")
root = root.rightnode
if __name__ == '__main__':
a = binarysearch()
a.insert(7,7)
a.insert(1,1)
a.insert(8,8)
a.insert(3,3)
a.insert(9,9)
a.insert(2,2)
a.insert(4,4)
a.insert(11,11)
a.insert(10,10)
a.deletemin()
a.getnodes()
The issue you have is that root = root.rightnode only rebinds the local variable root. It doesn't change the other places you have references to that node (such as its parent in the tree).
To fix this, you need to change how your recursive function works. Rather than expecting it to do all the work in the last call, it should instead return the value that should be the left node of its parent. Of then that will be the node itself, but for the minimum node, it will be its right child instead.
def deletemin(self):
if self.rootnode is None:
print("No nodes exist")
else:
self.rootnode = self.deleteminnode(self.rootnode)
def deleteminnode(self, root):
if root.leftnode is not None:
root.leftnode = self.deleteminnode(root.leftnode)
return root
else:
return root.rightnode
A final note regarding names: It's a bit weird to use root as the name of a random node within the tree. Usually a tree has just the one root node, and others nodes aren't called root since they have parents. Unfortunately, the most conventional name node is already being used for your node class. Normally classes should be given CapitalizedNames, so that lowercase_names can exclusively refer to instances and other variables. This is just convention though (and builtin types like list break the rules). It might be easier for others to understand your code if you use standard name styles, but Python doesn't enforce them. It will allow you to use whatever names you want. Even the name self is not a requirement, though it would be very confusing if you used something different for the first argument of a method without a good reason (an example of a good reason: classmethods and methods of metaclasses often use cls as the name of their first arguments, since the object will be a class).
You can find all the nodes in the tree, along with the path to the node, find the minimum of the results, and then traverse the generated path to delete the node:
class Tree:
def __init__(self, **kwargs):
self.__dict__ = {i:kwargs.get(i) for i in ['val', 'left', 'right']}
def get_nodes(self, current = []):
yield [''.join(current), self.val]
yield from getattr(self.right, 'get_nodes', lambda _:[])(current+['1'])
yield from getattr(self.left, 'get_nodes', lambda _:[])(current+['0'])
def __iter__(self):
yield self.val
yield from [[], self.left][bool(self.left)]
yield from [[], self.right][bool(self.right)]
def _insert_back(self, _v):
if not self.val:
self.val = _v
else:
if _v < self.val:
getattr(self.left, '_insert_back', lambda x:setattr(x, 'left', Tree(val=x)))(_v)
else:
getattr(self.right, '_insert_back', lambda x:setattr(x, 'right', Tree(val=x)))(_v)
def remove(self, _path, _to_val, last=None):
'''_to_val: if _to_val is None, then the item is removed. If not, the node value is set to _to_val'''
if _path:
getattr(self, ['left', 'right'][int(_path[0])]).remove(_path[1:], _to_val, last = self)
else:
if _to_val is None:
last.left = None
last.right = None
for i in [[], self.left][bool(self.left)]:
last._insert_back(i)
for i in [[], self.right][bool(self.right)]:
last._insert_back(i)
else:
self.val = _to_val
Creating:
7
5 9
4 6 8 10
12
t = Tree(val = 7, left=Tree(val = 5, left=Tree(val=4), right=Tree(val=6)), right=Tree(val=9, left=Tree(val=8), right=Tree(val=10, right=Tree(val=12))))
path, _to_remove = min(t.get_nodes(), key=lambda x:x[-1])
print(f'Removing {_to_remove}')
t.remove(path, None)
print([i for i in t])
Output:
4
[7, 5, 9, 8, 10, 12]

Python BST not working

I'm new to Python thus the question,this is the implementation of my my BST
class BST(object):
def __init__(self):
self.root = None
self.size = 0
def add(self, item):
return self.addHelper(item, self.root)
def addHelper(self, item, root):
if root is None:
root = Node(item)
return root
if item < root.data:
root.left = self.addHelper(item, root.left)
else:
root.right = self.addHelper(item, root.right)
This is the Node object
class Node(object):
def __init__(self, data):
self.data = data
self.left = None
self.right = None
This is my implmentation of str
def __str__(self):
self.levelByLevel(self.root)
return "Complete"
def levelByLevel(self, root):
delim = Node(sys.maxsize)
queue = deque()
queue.append(root)
queue.append(delim)
while queue:
temp = queue.popleft()
if temp == delim and len(queue) > 0:
queue.append(delim)
print()
else:
print(temp.data, " ")
if temp.left:
queue.append(temp.left)
if temp.right:
queue.append(temp.right)
This is my calling client,
def main():
bst = BST()
bst.root = bst.add(12)
bst.root = bst.add(15)
bst.root = bst.add(9)
bst.levelByLevel(bst.root)
if __name__ == '__main__':
main()
Instead of the expected output of printing the BST level by level I get the following output,
9
9223372036854775807
When I look in the debugger it seems that the every time the add method is called it starts with root as None and then returns the last number as root. I'm not sure why this is happening.
Any help appreciated.
If the root argument of your addHelper is None, you set it to a newly-created Node object and return it. If it is not, then you modify the argument but return nothing, so you end up setting bst.root to None again. Try the following with your code above — it should help your understanding of what your code is doing.
bst = BST()
bst.root = bst.add(12)
try:
print(bst.root.data)
except AttributeError:
print('root is None')
# => 12
# `bst.addHelper(12, self.root)` returned `Node(12)`,
# which `bst.add` returned too, so now `bst.root`
# is `Node(12)`
bst.root = bst.add(15)
try:
print(bst.root.data)
except AttributeError:
print('root is None')
# => root is None
# `bst.addHelper(15, self.root)` returned `None`,
# which `bst.add` returned too, so now `bst.root`
# is `None`.
bst.root = bst.add(9)
try:
print(bst.root.data)
except AttributeError:
print('root is None')
# => 9
# `bst.addHelper(9, self.root)` returned `Node(9)`,
# which `bst.add` returned too, so now `bst.root`
# is `Node(9)`
So you should do two things:
make you addHelper always return its last argument — after the appropriate modifications —, and
have your add function take care of assigning the result to self.root (do not leave it for the class user to do).
Here is the code:
def add(self, item):
self.root = self.addHelper(item, self.root)
self.size += 1 # Otherwise what good is `self.size`?
def addHelper(self, item, node):
if node is None:
node = Node(item)
elif item < node.data:
node.left = self.addHelper(item, node.left)
else:
node.right = self.addHelper(item, node.right)
return node
Notice that I changed the name of the last argument in addHelper to node for clarity (there already is something called root: that of the tree!).
You can now write your main function as follows:
def main():
bst = BST()
bst.add(12)
bst.add(15)
bst.add(9)
bst.levelByLevel(bst.root)
(which is exactly what #AaronTaggart suggests — but you need the modifications in add and addHelper). Its output is:
12
9
15
9223372036854775807
The above gets you to a working binary search tree. A few notes:
I would further modify your levelByLevel to avoid printing that last value, as well as not taking any arguments (besides self, of course) — it should always print from the root of the tree.
bst.add(None) will raise an error. You can guard against it by changing your add method. One possibility is
def add(self, item):
try:
self.root = self.addHelper(item, self.root)
self.size += 1
except TypeError:
pass
Another option (faster, since it refuses to go on processing item if it is None) is
def add(self, item):
if item is not None:
self.root = self.addHelper(item, self.root)
self.size += 1
From the point of view of design, I would expect selecting a node from a binary search tree would give me the subtree below it. In a way it does (the node contains references to all other nodes below), but still: Node and BST objects are different things. You may want to think about a way of unifying the two (this is the point in #YairTwito's answer).
One last thing: in Python, the convention for naming things is to have words in lower case and separated by underscores, not the camelCasing you are using — so add_helper instead of addHelper. I would further add an underscore at the beginning to signal that it is not meant for public use — so _add_helper, or simply _add.
Based on the following, you can see that bst.root in None after the second call to add():
>>> bst.root = bst.add(12)
>>> bst.root
<__main__.Node object at 0x7f9aaa29cfd0>
>>> bst.root = bst.add(15)
>>> type(bst.root)
<type 'NoneType'>
Your addHelper isn't returning the root node. Try this:
def addHelper(self, item, root):
if root is None:
root = Node(item)
return root
if item < root.data:
root.left = self.addHelper(item, root.left)
else:
root.right = self.addHelper(item, root.right)
return root
And then it works as expected:
>>> bst.root = bst.add(12)
>>> bst.root = bst.add(15)
>>> bst.levelByLevel(bst.root)
(12, ' ')
()
(15, ' ')
(9223372036854775807, ' ')
>>> bst.root = bst.add(9)
>>> bst.levelByLevel(bst.root)
(12, ' ')
()
(9, ' ')
(15, ' ')
(9223372036854775807, ' ')
You're using the BST object basically only to hold a root Node and the add function doesn't really operate on the BST object so it's better to have only one class (BtsNode) and implement the add there. Try that and you'll see that the add function would be much simpler.
And, in general, when a member function doesn't use self it shouldn't be a member function (like addHelper), i.e., it shouldn't have self as a parameter (if you'd like I can show you how to write the BtsNode class).
I tried writing a class that uses your idea of how to implement the BST.
class BstNode:
def __init__(self):
self.left = None
self.right = None
self.data = None
def add(self,item):
if not self.data:
self.data = item
elif item >= self.data:
if not self.right:
self.right = BstNode()
self.right.add(item)
else:
if not self.left:
self.left = BstNode()
self.left.add(item)
That way you can create a BST the following way:
bst = BstNode()
bst.add(13)
bst.add(10)
bst.add(20)
The difference is that now the add function actually operates on the object without any need for the user to do anything. The function changes the state of the object by itself.
In general a function should do only what it's expected to do. The add function is expected to add an item to the tree so it shouldn't return the root. The fact that you had to write bst.root = bst.add() each time should signal that there's some fault in your design.
Your add method probably shouldn't return a value. And you most certainly shouldn't assign the root of the tree to what the add method returns.
Try changing your main code to something like this:
def main():
bst = BST()
bst.add(12)
bst.add(15)
bst.add(9)
bst.levelByLevel(bst.root)
if __name__ == '__main__':
main()

How to create new class instances dynamically at runtime in python?

I am trying to solve this problem:
Imagine a (literal) stack of plates. If the stack gets too high, it
might topple. There- fore, in real life, we would likely start a new
stack when the previous stack exceeds some threshold. Implement a data
structure SetOfStacks that mimics this. SetOf- Stacks should be
composed of several stacks, and should create a new stack once the
previous one exceeds capacity. SetOfStacks.push() and
SetOfStacks.pop() should behave identically to a single stack (that
is, pop() should return the same values as it would if there were just
a single stack). Bonus: Implement a function popAt(int index) which
performs a pop operation on a specific sub-stack.
So I wrote the code:
#!/bin/env python
from types import *
class Stack:
def __init__(self):
self.items = []
self.capacity = 3
self.stackscount = 0
def create(self):
id = self.stackscount + 1
id = str(id) + "_stack"
# How to create a new instance of Stack class at runtime ?
# the __init__ must be run too.
def push(self, item):
if self.size() <= self.capacity:
self.items.append(item)
else:
self.create()
def pop(self):
return self.items.pop()
def popAt(self):
pass
def peek(self):
return self.items[len(self.items)-1]
def size(self):
return len(self.items)
s = Stack()
s.push(10)
How do I create a new s type object dynamically at runtime? I searched on the internet and found that using new.instance or new.classobj is the solution but when I did so my new object did not seem to have items from __init__ function. In python3, type() seems to be the answer but the docs doesn't have any examples.
You've confused yourself by referring to a "type object". In Python that means the class itself, not its instances.
To create new Stack objects, simply do what you're already doing: call the Stack class. You can append them to a list:
stacks = [Stack() for _ in range(5)]
However, as jon points out, that won't solve your problem since you haven't defined the SetOfStacks class.
You could simply use a parent-child relation : when a Stack is full, it creates a child and delegate next pushes to it. It could lead to :
class Stack:
def __init__(self, parent = None, id=None):
self.stackscount = 0
self.capacity = 3
self.items = []
self.parent = parent
self.id = id
self.child = None
def create(self):
id = self.stackscount + 1
id = str(id) + "_stack"
return Stack(self, id)
def push(self, item):
if self.size() <= self.capacity:
self.items.append(item)
else:
if self.child is None:
self.child = self.create()
self.child.push(item)
def pop(self):
if self.child is not None:
item = self.child.pop()
if len(self.child.items) == 0:
self.child = None
else:
item = self.items.pop()
return item
def popAt(self):
pass
def peek(self):
if self.child is not None:
item = self.child.peek()
else:
item = self.items[len(self.items)-1]
return item
def size(self):
l = len(self.items)
if self.child is not None:
l += self.child.size()
return l
s = Stack()
s.push(10)
popAt is still to be implemented, but I tested it and it correctly creates new stacks when pushing and empties and removes them when popping.
The implementation of popAt will require some evolutions to current pop implementation, to allow removing an intermediate stack :
def pop(self):
if self.child is not None:
item = self.child.pop()
if len(self.child.items) == 0:
self.child = self.child.child
if self.child is not None:
self.child.parent = self
else:
item = self.items.pop()
return item
def popAt(self, stacknumber):
s = self
for i in range(stacknumber):
s = s.child
if s is None:
return None
if len(s.items) == 0:
return None
item = s.items.pop()
if len(s.items) == 0 and s.parent is not None:
s.parent.child = s.child
if s.child is not None:
s.child.parent = s.parent
return item
The type() function is indeed what you are looking for. Documentation can be found here: https://docs.python.org/2/library/functions.html#type
You can call it like this:
# Bases is a tuple of parent classes to inherit
bases = Stack,
# Dict contains extra properties for the class, for example if you want to add a class variable or function
dict_ = {}
# Construct the class
YourClass = type('YourClass', bases, dict_)
# Create an instance of the class
your_instance = YourClass()
It looks like you are just looking at instance creation though:
class Stack(object):
def create(self):
id = self.stackscount + 1
id = str(id) + "_stack"
# How to create a new instance of Stack class at runtime ?
# the __init__ must be run too.
stack = Stack()

Binary Search Tree operations

I am trying to implement Binary Search Tree operations in python. As of now, I have written some code to add nodes to this search tree (sorted).
Here's what I've in my code:
class TreeNode:
def __init__(self, data):
self.data = data
self.lLink = None
self.rLink = None
class BinaryTree:
def __init__(self):
self.root = None
def AddNode(self, data):
if self.root is None:
self.root = TreeNode(data)
else:
if data < self.root.data:
if self.root.lLink is None:
self.root.lLink = TreeNode(data)
else:
AddNode(self.root.lLink, data)
else:
if self.root.rLink is None:
self.root.rLink = TreeNode(data)
else:
AddNode(self.root.rLink, data)
def InOrder(self, head):
if self.root.lLink is not None:
InOrder(self.root.lLink)
print self.root.data,
if self.root.rLink is not None:
InOrder(self.root.rLink)
myTree = BinaryTree()
myTree.AddNode(15)
myTree.AddNode(18)
myTree.AddNode(14)
How do I test if my AddNode() method is correct? I know the algorithm but just to be sure.
What I was thinking of is to create an InOrder() method and try to print elements through this InOrder traversal. As a result, my data added to the tree should be displayed in sorted order. If it is displayed in sorted order, I'll be sure that both my AddNode() and InOrder() methods are correct.
Your BinaryTree class is faulty, changing the order of insertions to
myTree.AddNode(14)
myTree.AddNode(18)
myTree.AddNode(15)
raises an error - NameError: global name 'AddNode' is not defined.
This is because in the lines, AddNode(self.root.rLink, data) and AddNode(self.root.lLink, data) you seem to be calling the AddNode function on instances of TreeNode which is not possible. I fixed up some of the errors in your code and it should work great now.
class TreeNode:
def __init__(self, data):
self.data = data
self.lLink = None
self.rLink = None
class BinaryTree:
def __init__(self):
self.root = None
def AddNode(self, data):
if self.root is None:
self.root = TreeNode(data)
else:
self.AddHelper(data, self.root)
def AddHelper(self, data, startingPoint):
if data < startingPoint.data:
if startingPoint.lLink is None:
startingPoint.lLink = TreeNode(data)
else:
self.AddHelper(data, startingPoint.lLink)
else:
if startingPoint.rLink is None:
startingPoint.rLink = TreeNode(data)
else:
self.AddHelper(data, startingPoint.rLink)
def InOrder(self):
self.InOrderHelper(self.root)
def InOrderHelper(self, startingPoint):
if startingPoint is None:
return
self.InOrderHelper(startingPoint.lLink)
print startingPoint.data,
self.InOrderHelper(startingPoint.rLink)
Output Test :
>>> myTree = BinaryTree()
>>> myTree.AddNode(14)
>>> myTree.AddNode(18)
>>> myTree.AddNode(15)
>>> myTree.InOrder()
14 15 18
Inserting can be a little tricky, especially because the function is a part of the tree itself. So, you call the insert function on the tree, but specifying a starting point. This defaults to root, so you can leave the argument when you call the function.
Also, I think you are a little unclear about how self works in a function. You cannot pass it as an argument to the function, which is what it seems you have done.
class TreeNode:
def __init__(self, data):
self.data = data
self.rLink = None
self.lLink = None
class BinaryTree:
def __init__(self):
self.root = None
def AddNode(self, data, node=None):
if not node :
node = self.root
if self.root is None:
self.root = TreeNode(data)
else:
if data < node.data:
if node.lLink is None:
node.lLink = TreeNode(data)
else:
self.AddNode(data, self.root.lLink)
else:
if node.rLink is None:
node.rLink = TreeNode(data)
else:
self.AddNode(data, self.root.rLink)
def InOrder(self, head):
if head.lLink is not None:
self.InOrder(head.lLink)
print head.data,
if head.rLink is not None:
self.InOrder(head.rLink)
myTree = BinaryTree()
myTree.AddNode(14)
myTree.AddNode(15)
myTree.AddNode(18)
myTree.InOrder(myTree.root)
Testing the insert function with an in-order traversal is the best approach.
This should work. You are not going down the tree if you use self.root.lLink every time.
Optionally, you could write one more line of code to check if the output is indeed in ascending order.

Creating a List from a Binary Search Tree

I'm trying to make a list of all items in a binary search tree. I understand the recursion but I don't know how to make it return each value and then append it into a list. I want to create a function called makeList() that will return a list of all the items in my tree. All the functions in my programs work except the makeList() function and are included to make sure everyone understands the basic structure of how I set up my tree.
class Node(object):
def __init__(self, data):
self.data = data
self.lChild = None
self.rChild = None
class Tree(object):
def __init__(self):
self.root = None
def __str__(self):
current = self.root
def isEmpty(self):
if self.root == None:
return True
else:
return False
def insert (self, item):
newNode = Node (item)
current = self.root
parent = self.root
if self.root == None:
self.root = newNode
else:
while current != None:
parent = current
if item < current.data:
current = current.lChild
else:
current = current.rChild
if item < parent.data:
parent.lChild = newNode
else:
parent.rChild = newNode
def inOrder(self, aNode):
if aNode == None:
pass
if aNode != None:
self.inOrder(aNode.lChild)
print aNode.data
self.inOrder(aNode.rChild)
def makeList(self, aNode):
a = []
self.inOrder(aNode)
a += [aNode.data]
print a
n = Tree()
for i in [4,7,2,9,1]:
n.insert(i)
n.makeList(n.root)
Looking at my makeList() function I can see why it doesn't work but I don't know how to make it work.
EDIT
Ok, I got it! And I even got two answers which are:
def makeList(self, aNode, a = []):
if aNode != None:
self.makeList(aNode.lChild, a)
a += [aNode.data]
self.makeList(aNode.rChild, a)
return a
and
def makeList2(self, aNode):
if aNode is None:
return []
return self.makeList2(aNode.lChild) + [aNode.data] + self.makeList2(aNode.rChild)
And looking back I can see that I do not understand recursion very well so it's time to hit the books! Anyone have any good resources on recursion?
Another question, so say I call my makeList() function. When Python goes through makeList(), when it gets to the self.makeList(aNode.lChild, a) does it begin running the function again while it's still finishing up the makeList() function or does everything stop and it just starts over with it's new aNode?
I hope that makes sense.
You're so close! makeList can be pretty simple:
def makeList(self, aNode):
if aNode is None:
# Stop recursing here
return []
return self.makeList(aNode.lChild) + [aNode.data] + self.makeList(aNode.rChild)
Basically, make sure you're not trying to recurse past empty nodes. Then return the list of the left tree, the current node, and the list of the right tree.
inOrder prints things but does not return anything, so it's useless for building a list. You need a way to return each node in order. This may be something that your class hasn't covered yet, but check out the yield command.
The basic idea is something like this:
def makeList(self):
return self.lChild.makeList() + [self.data] + self.rChild.makeList()
See how it is essentially the same thing as inOrder?
You have a different structure in your program that makes it a bit harder to implement, but the basic idea is the same.

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