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]
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 create a binary tree using a linked list. I want to delete a node in a binary tree, the way I implemented is to set the address(pointer) to the branch that I want to delete as None, but when I run the traversal methods, the branch still shows up.
here is my code
class tree:
def __init__(self,val):
self.val=val
self.left=None
self.right=None
def preorder(root):
if root is None:
return
print(root.val)
tree.preorder(root.left)
tree.preorder(root.right)
def inorder(root):
if root is None:
return
tree.inorder(root.left)
print(root.val)
tree.inorder(root.right)
def postorder(root):
if root is None:
return
tree.postorder(root.left)
tree.postorder(root.right)
print(root.val)
def levelorder(root):
if root is None:
return
Q=[]
Q.append(root)
while Q!=[]:
l=len(Q)
for i in range(l):
print(Q[i].val)
temp=[]
for i in Q:
if i.left is not None:
temp.append(i.left)
if i.right is not None:
temp.append(i.right)
Q.clear()
Q=temp.copy()
def search(root,value):
o=[]
o.append(tree.levelorder(root))
if value in o:
return "Found"
else:
return "Not Found"
def insert(root,value,where):
if root is None:
return
Q=[]
Q.append(root)
value=tree(value)
while Q!=[]:
for i in Q:
if i.val==where:
if i.left is not None:
i.right=value
else:
i.left=value
return
temp=[]
for i in Q:
if i.left is not None:
temp.append(i.left)
if i.right is not None:
temp.append(i.right)
Q.clear()
Q=temp.copy()
def delete(root,value):
if root is None:
return
Q=[]
Q.append(root)
value=tree(value)
while Q!=[]:
for i in Q:
if i.left is not None and i.left.val==value:
i.left.val=None
i.left=None
if i.right is not None and i.right.val==value:
i.right.val=None
i.right=None
return
temp=[]
for i in Q:
if i.left is not None:
temp.append(i.left)
if i.right is not None:
temp.append(i.right)
Q.clear()
Q=temp.copy()
and here is how I created the tree
base=tree("drinks")
L=tree("hot")
R=tree("cold")
LL=tree("coffe")
LR=tree("tea")
LRL=tree("w milk")
LRR=tree("wo milk")
base.left=L
base.right=R
L.left=LL
L.right=LR
LR.left=LRL
LR.right=LRR
Now when I run the delete method, the object that is supposed to be deleted still shows up.
tree.delete(base,"w milk")
tree.levelorder(base)
drinks
hot
cold
coffe
tea
ice cream
w milk
wo milk <=== the node i am trying to delete
The main issues:
delete method's while loop will exit immediately, because of the return statement
The value to compare with should not be turned into a node with value=tree(value)
Some remarks on your code:
You'll not be able to ever delete the root node of the tree, since you only check the values of child nodes. In order to allow the root to be deleted, and to represent an empty tree, you should really consider creating a second class. The current class could then be renamed to Node and the new class could become Tree.
It is a common habit to use a capital first letter for class names, while for variable names you would always start with a lower case letter. So not Q, but q.
It is not such good practice to print inside methods of such classes: these methods should just provide tools to iterate over nodes, but they should not print them. Use yield instead.
For instance, search should return the node when it is found and None otherwise. Don't print the result.
The insert method may actually delete node(s) when both the left and right child of the where node are already occupied. It would be better to reject such an insert.
In the OOP pattern, it is widely accepted to call the first argument of instance methods self. In OOP, you would call these methods with the dot notation: so instead of tree.postorder(root.left) you would do root.left.postorder().
You have quite some code repetition: mainly for traversals, and where you use the queue algorithm twice. Try to avoid that.
Here is how you could modify your code to align to these suggestions:
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def traverse(self, kind=0, parent=None):
if kind == 0: # pre-order
yield (self, parent) # don't print in methods
if self.left:
yield from self.left.traverse(kind, self)
if kind == 1: # in-order
yield (self, parent)
if self.right:
yield from self.right.traverse(kind, self)
if kind == 2: # post-order
yield (self, parent)
class Tree:
def __init__(self):
self.root = None # Representing an empty tree
def traversevalues(self, kind=0):
if self.root:
for node, parent in self.root.traverse(kind):
yield node.val
def preorder(self):
yield from self.traversevalues(0)
def inorder(self):
yield from self.traversevalues(1)
def postorder(self):
yield from self.traversevalues(2)
def levelorder(self):
if self.root:
q = [self.root]
while q:
temp = []
for node in q:
yield node.val
if node.left:
temp.append(node.left)
if node.right:
temp.append(node.right)
q = temp
def search(self, value):
if self.root:
return next((edge for edge in self.root.traverse() if edge[0].val == value), None)
def insert(self, value, where=None):
if not where:
if self.root:
raise ValueError("Must call insert with second argument")
else:
self.root = Node(value)
else:
edge = self.search(where)
if edge:
parent = edge[0]
if not parent.left:
parent.left = Node(value)
elif not parent.right:
parent.right = Node(value)
else:
raise ValueError(f"Node {where} already has 2 children")
else:
raise ValueError(f"Node {where} not found")
def deletesubtree(self, value):
edge = self.search(value)
if edge:
node, parent = edge
if parent.left == node:
parent.left = None
else:
parent.right = None
else:
raise ValueError(f"Node {value} not found")
tree = Tree()
tree.insert("drinks")
tree.insert("hot", "drinks")
tree.insert("cold", "drinks")
tree.insert("coffee", "hot")
tree.insert("tea", "hot")
tree.insert("with milk", "tea")
tree.insert("without milk", "tea")
print(*tree.levelorder())
tree.deletesubtree("without milk")
print(*tree.levelorder())
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 have been struggling with this problem for a while and I am a Python beginner when it comes to BST, so I would appreciate some help. I am dynamically adding elements from an (unsorted) array into BST. That part is fine, I know how to do that. The next step, proved to be impossible with my current skill set. As I am adding elements to the tree, I need to be able to find current rank of any element in the tree. I know there are subtleties in this problem, so I would need help to at least find the number of nodes that are below the given node in BST. For example, in this case, node 15 has nodes 10,5 and 13 below it, so the function will return 3. Here is my existing code [this is a problem from Cracking the coding interview, chapter 11]
class Node:
"""docstring for Node"""
def __init__(self, data):
self.data = data
self.left=None
self.right=None
self.numLeftChildren=0
self.numRightChildren=0
class BSTree:
def __init__(self):
self.root = None
def addNode(self, data):
return Node(data)
def insert(self, root, data):
if root == None:
return self.addNode(data)
else:
if data <= root.data:
root.numLeftChildren+=1
root.left = self.insert(root.left, data)
else:
root.numRightChildren+=1
root.right = self.insert(root.right, data)
return root
def getRankOfNumber(self,root,x):
if root==None:
return 0
else:
if x>root.data :
return self.getRankOfNumber(root.right,x)+root.numLeftChildren+1
elif root.data==x:
return root.numLeftChildren
else:
return self.getRankOfNumber(root.left,x)
BTree=BSTree()
root=BTree.addNode(20)
BTree.insert(root,25)
BTree.insert(root,15)
BTree.insert(root,10)
BTree.insert(root,5)
BTree.insert(root,13)
BTree.insert(root,23)
You can go by this approach:
1. Have 2 more fields in each node numLeftChildren and numRightChildren.
2. Initialize both of them to 0 when you create a new node.
3. At the time of insertion, you make a comparison if the newly added node's
key is less than root's key than you increment, root's numLeftChildren and
call recursion on root's left child with the new node.
4. Do Same thing if new node's key is greater than root's key.
Now, come back to your original problem, You have to find out the number of children in left subtree. Just find out that node in O(logN) time and just print the numLeftChildren
Time Complexity: O(logN)
PS: I have added another field numRightChildren which you can remove if you are always interested in knowing the number of nodes in left subtree only.
You could modify your BST to contain the number of nodes beneath each node.
Or you could iterate over a traditional BST from least to greatest, counting as you go, and stop counting when you find a node of the required value.
You could simplify the code by using just instances of the Node class, as your BSTree instance operates on the root node anyway. Another optimization could be to not represent duplicate values as separate nodes, but instead use a counter of occurrences of the key:
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
self.num_left_children = 0
self.occurrences = 1
def insert(self, data):
if data < self.data:
if self.left is None:
self.left = Node(data)
else:
self.left.insert(data)
self.num_left_children += 1
elif data > self.data:
if self.right is None:
self.right = Node(data)
else:
self.right.insert(data)
else:
self.occurrences += 1
def get_rank(self, data):
if data < self.data:
return self.left.get_rank(data)
elif data > self.data:
return (self.occurrences + self.num_left_children +
self.right.get_rank(data))
else:
return self.num_left_children
Demo:
root = Node(20)
root.insert(25)
root.insert(15)
root.insert(10)
root.insert(10)
root.insert(5)
root.insert(13)
root.insert(23)
print(root.get_rank(15)) # 4
You can use the below function to find the number of the nodes in the tree (including the root).
def countNodes(self, root):
if root == None:
return 0
else
return (1 + countNodes(root.left) + countNodes(root.right));
To find the number of nodes that lie below root, subtract 1 from the value returned by the function. I think this will help get you started on the problem.
Your code will look like:
class Node:
"""docstring for Node"""
def init(self, data):
self.data = data
self.left=None
self.right=None
self.depth=0
class BSTree:
def init(self):
self.root = None
def addNode(self, data):
return Node(data)
def insert(self, root, data):
if root == None:
return self.addNode(data)
else:
if data <= root.data:
root.left = self.insert(root.left, data)
else:
root.right = self.insert(root.right, data)
return root
BTree=BSTree()
root=BTree.addNode(20)
BTree.insert(root,25)
BTree.insert(root,15)
BTree.insert(root,10)
BTree.insert(root,5)
BTree.insert(root,13)
BTree.insert(root,23)
BTree.insert(root,23)
numLeft = countNodes(root->left);
numRight = countNodes(root->right);
numChildren = numLeft + numRight;