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.开发者_如何转开发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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