I'm trying to use recursion to traverse a binary tree. Each tree either has two children, or it has no children (that is, the fields reserved for children == None)
I'd like to add the final leaves of each branch (that is, each Node whose two children == None) to a list, and return the list. I'm doing this with the 'search' function, and the helper 'search_base' function.
Through the debugger, I see that the list within the 'search' function indeed contains the elements I want it to. But, when it's returned in the search_base function, the result seems to be an empty list.
I'm extremely confused and would be grateful for any help. Thank you!
class Node:
def __init__(self, data, pos = None, neg = None):
self.data = data
self.positive_child = pos
self.negative_child = neg
class Diagnoser:
def __init__(self, root):
self.root = root
def search_base(self):
leaf_list=[]
current = self.root
return self.search(current, leaf_list)
def search(self, current, leaf_list):
if(current.positive_child == None):
leaf_list.append(current)
return leaf_list
else:
self.search(current.positive_child, leaf_list)
self.search(current.negative_child, leaf_list)
if __name__ == "__main__":
# Manually build a simple tree.
# cough
# Yes / \ No
# fever healthy
# Yes / \ No
# influenza cold
flu_leaf = Node("influenza", None, None)
cold_leaf = Node("cold", None, None)
inner_vertex = Node("fever", flu_leaf, cold_leaf)
healthy_leaf = Node("healthy", None, None)
root = Node("cough", inner_vertex, healthy_leaf)
diagnoser = Diagnoser(root)
leaf_list = diagnoser.search_base()
print(leaf_list[0].data)
The problem is that in
self.search(current.positive_child, leaf_list)
self.search(current.negative_child, leaf_list)
The return values from these statements is not saved or returned, so here the function gives None. Also, the leaf_list passed into both these statements is the same, i.e., they don't get concatenated. In recursive functions you it's best not to have side effects to keep it safe.
It should be:
def search(self, current, leaf_list=[]):
if(current.positive_child == None):
return [current]
else:
return (self.search(current.positive_child, leaf_list)
+ self.search(current.negative_child, leaf_list))
Since search modifies the list, it doesn't need to return anything, and search_base can just return the modified list.
class Diagnoser:
def __init__(self, root):
self.root = root
def search_base(self):
leaf_list = []
current = self.root
self.search(current, leaf_list)
return leaf_list
def search(self, current, leaf_list):
if current.positive_child is None:
leaf_list.append(current)
else:
self.search(current.positive_child, leaf_list)
self.search(current.negative_child, leaf_list)
Also, you need to check that both children are missing, i.e.
if current.positive_child is None and current.negative_child is None:
Here's a full simpler solution without side effects:
class Node:
def __init__(self, data, pos=None, neg=None):
self.data = data
self.positive_child = pos
self.negative_child = neg
def leaves(self):
if self.positive_child is self.negative_child is None:
return [self]
else:
return (self.positive_child.leaves() +
self.negative_child.leaves())
if __name__ == "__main__":
# Manually build a simple tree.
# cough
# Yes / \ No
# fever healthy
# Yes / \ No
# influenza cold
flu_leaf = Node("influenza", None, None)
cold_leaf = Node("cold", None, None)
inner_vertex = Node("fever", flu_leaf, cold_leaf)
healthy_leaf = Node("healthy", None, None)
root = Node("cough", inner_vertex, healthy_leaf)
for leaf in root.leaves():
print(leaf.data)
Related
I have below method where self contains a data structure as below
self.place = "India"
self.children = ["Tamil Nadu", "Karnataka"]
self.parent
Method
def get_node(self, value):
if value is None:
return self
if self.place == value:
return self
for node in self.children:
if node.place == value:
return node
elif len(node.children) > 0:
return node.get_node(value)
So via recursion, I am iterating on all possible child nodes to find the node I am looking for via return node.get_node(value) but I observed that, iteration happening via "Tamil Nadu" but not via "Karnataka".
I understood that, it took the first element of the list and then continued from there, but not coming back to 2nd element of the list.
is this expected behavior from recursion or am I doing something wrong ?
Full code( In case needed for testing)
class TreeNode:
def __init__(self, place):
self.place = place
self.children = []
self.parent = None
def add_child(self, child):
child.parent = self
self.children.append(child)
def print_tree(self):
prefix = ""
if self.parent is None:
print(self.place)
else:
prefix = prefix + (" " * self.get_level() * 3)
prefix = prefix + "|__"
print(prefix + self.place)
for child in self.children:
child.print_tree()
def get_level(self):
level = 0
p = self.parent
while p:
level = level + 1
p = p.parent
return level
def get_node(self, value):
if value is None:
return self
if self.place == value:
return self
for node in self.children:
if node.place == value:
return node
elif len(node.children) > 0:
return node.get_node(value)
def tree_map(self, nodes):
for node in nodes:
self.add_child(TreeNode(node))
def build_places():
root = TreeNode("Global")
india = TreeNode("India")
usa = TreeNode("USA")
root.add_child(india)
root.add_child(usa)
india_nodes = ["Gujarat" ,"Karnataka"]
gujarath_nodes = [ "Ahmedabad", "Baroda"]
karnataka_nodes = ["Bangalore", "Mysore"]
usa_nodes = ["New Jersey", "California"]
newjersey_nodes = ["Princeton", "Trenton"]
california_nodes = ["San Franciso", "Mountain View", "Palo Alto"]
for node in india_nodes:
india.add_child(TreeNode(node))
for node in usa_nodes:
usa.add_child(TreeNode(node))
gujarath_node = root.get_node("Gujarat")
print(gujarath_node.place)
for node in gujarath_nodes:
gujarath_node.add_child(TreeNode(node))
karnataka_node = root.get_node("Karnataka")
print(karnataka_node.place)
return root
if __name__ == "__main__":
root = build_places()
root.print_tree()
The problem is that in your loop you are always exiting the loop in its first iteration (when the node has at least some children). You should only exit on success, not when the recursive call comes back without success.
So change the loop to this:
for node in self.children:
if node.place == value:
return node
elif len(node.children) > 0:
result = node.get_node(value)
if result:
return result
Secondly, there is a strange base case you have at the start of this function. I would replace this:
if value is None:
return self
With:
if value is None:
return None
...since you didn't look for the value in that case: so then (in my opinion) it is not right to return a node instance (which might have any value -- you didn't verify it). It seems more consistent to return None or to remove this whole if block and not treat None in a special way.
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]
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
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()
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.