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Trouble balancing a binary search tree using AVL

I'm unable to identify where I'm going wrong with my AVL implementation for balancing an existing binary search tree. I'm not getting any errors but my binary search tree does not come out to be properly balanced. After insertion, my binary search tree looks like (it would be prudent here to mention that my display_keys method gives us a visualization that is rotated by 90 degrees):
∅
The-Dreamers
∅
Saint-Laurent
∅
Pierrot-le-Fou
∅
Contempt
Cold-War
Before-Sunrise
∅
Basic-Instinct
∅
This is correct as it seems be to be following the rules for a BST.
But after calling the BalanceTree() method on my binary search tree, I seem to get:
∅
The-Dreamers
Saint-Laurent
Pierrot-le-Fou
Contempt
Cold-War
Before-Sunrise
Basic-Instinct
∅
which as you can see, is not Balanced. But wait, and here's the catch, if I call BalanceTree() again, the tree comes out to be perfectly balanced and isBSTBalanced returns True also. After the second call to BalanceTree(), our tree looks like:
∅
The-Dreamers
Saint-Laurent
Pierrot-le-Fou
Contempt
Cold-War
Before-Sunrise
Basic-Instinct
∅
I am adding the complete source code for my BST class for clarity and if somebody wants to execute the code, but I have added a comment (#Addition of new methods for AVL starts here) in the BST class to indicate where the methods for AVL start. You need only concern yourself with them. I would like for you help me pinpoint what exactly is going wrong in my code.
class BST:
class TreeNode:
def __init__(self, key, value, left=None, right=None, parent=None):
self.key = key
self.value = value
self.left = left
self.right = right
self.parent = parent
self.height = 1
def __init__(self):
self.root = None
self.size = 0
def __len__(self):
return self.size
def insert(self, key, value):
if self.root == None:
self.root = self.TreeNode(key, value)
else:
self._insert(key, value, self.root)
self.size += 1
def _insert(self, key, value, curr_node):
if key < curr_node.key:
if curr_node.left is not None:
self._insert(key, value, curr_node.left)
else:
curr_node.left = self.TreeNode(key, value, parent=curr_node)
elif key > curr_node.key:
if curr_node.right is not None:
self._insert(key, value, curr_node.right)
else:
curr_node.right = self.TreeNode(key, value, parent=curr_node)
def search(self, key):
if self.root:
found = self._search(key, self.root)
if found:
return found.value
else:
return None
else:
return None
def _search(self, key, curr_node):
if not curr_node:
return None
elif curr_node.key == key:
return curr_node
elif key < curr_node.key:
return self._search(key, curr_node.left)
else:
return self._search(key, curr_node.right)
def find_min(self):
curr = self.root
while curr.left is not None:
curr = curr.left
return curr
def find(self, node):
curr = node
while curr.left is not None:
curr = curr.left
return curr
def delete(self, key):
node_to_remove = self._search(key, self.root)
if node_to_remove.left is None and node_to_remove.right is None:
#Then we identify this as a leaf node
if node_to_remove is node_to_remove.parent.left:
#Setting the parent's reference to this to None
node_to_remove.parent.left = None
elif node_to_remove is node_to_remove.parent.right:
node_to_remove.parent.right = None
#2nd Case --> Two child
elif node_to_remove.left and node_to_remove.right:
minimum = self.find(node_to_remove.right)
self.delete(minimum.key) #We will still have a ref to this node afterwards
node_to_remove.key, node_to_remove.value = minimum.key, minimum.value
#3rd Case -> One child
else:
if node_to_remove.left:
node_to_remove.left.parent = node_to_remove.parent
node_to_remove.parent.left = node_to_remove.left
elif node_to_remove.right:
node_to_remove.right.parent = node_to_remove.parent
node_to_remove.parent.right = node_to_remove.right
def traversal(self, root):
res = []
if root:
res = self.traversal(root.left)
res.append(root)
res = res + self.traversal(root.right)
return res
def inorder_traversal(self, root):
if root:
self.inorder_traversal(root.left)
print(root.key)
self.inorder_traversal(root.right)
#Addition of new methods for AVL starts here
def display_keys(self, node, space='\t', level=0):
"""
Allows us to visualize the tree (albiet rotated by 90 degrees)
"""
# print(node.key if node else None, level)
# If the node is empty
if node is None:
print(space*level + '∅')
return
# If the node is a leaf
if node.left is None and node.right is None:
print(space*level + str(node.key))
return
# If the node has children
self.display_keys(node.right, space, level+1)
print(space*level + str(node.key))
self.display_keys(node.left,space, level+1)
def height(self):
return self._height(self.root)
def _height(self, curr_node):
if curr_node is None:
return -1 #since we are counting number of edges, we will return -1
else:
return 1 + max(self._height(curr_node.left), self._height(curr_node.right))
def isBSTBalanced(self):
return self._isBSTBalanced(self.root)
def _isBSTBalanced(self, curr_node):
if curr_node is None:
return True
hleft_subtree = self._height(curr_node.left)
hright_subtree = self._height(curr_node.right)
if hleft_subtree - hright_subtree in [-1,0,1]:
return self._isBSTBalanced(curr_node.left) and self._isBSTBalanced(curr_node.right)
else:
return False
def balance_factor(self):
if self.root is not None:
return self._balance_factor(self.root)
else:
return 0
def _balance_factor(self, curr_node):
if curr_node is None:
return
hleft_subtree = self._height(curr_node.left)
hright_subtree = self._height(curr_node.right)
b_factor = hleft_subtree - hright_subtree
return b_factor
def BalanceTree(self):
if self.isBSTBalanced() == False:
return self._rebalance(self.root)
def _rebalance(self, curr_node):
if curr_node is None:
return None
curr_node.left = self._rebalance(curr_node.left)
curr_node.right = self._rebalance(curr_node.right)
curr_node.height = 1 + max(self._height(curr_node.left), self._height(curr_node.right))
#print(curr_node.height)
if self._balance_factor(curr_node) > 1 and self._balance_factor(curr_node.left) >= 0:
#left heavy subtree
return self._rotate_right(curr_node)
if self._balance_factor(curr_node) < -1 and self._balance_factor(curr_node.right) <= 0:
#right heavy subtree
return self._rotate_left(curr_node)
if self._balance_factor(curr_node) < 0 and self._balance_factor(curr_node.right) > 0:
self._rotate_right(curr_node.right)
return self._rotate_left(curr_node)
if self._balance_factor(curr_node) > 0 and self._balance_factor(curr_node.left) < 0:
self._rotate_left(curr_node.left)
return self._rotate_right(curr_node)
return curr_node
def _rotate_left(self, oldRoot):
newRoot = oldRoot.right #the newRoot is the right child of the previous root
oldRoot.right = newRoot.left #replacing right child of the old root with the left child of the new
if newRoot.left is not None:
newRoot.left.parent = oldRoot
newRoot.parent = oldRoot.parent
if oldRoot == self.root:
self.root = newRoot
else:
if oldRoot.parent.left is oldRoot: #Checking isLeftChild
oldRoot.parent.left = newRoot
else:
oldRoot.parent.right = newRoot
newRoot.left = oldRoot
oldRoot.parent = newRoot
oldRoot.height = 1 + max(self._height(oldRoot.left), self._height(oldRoot.right))
newRoot.height = 1 + max(self._height(newRoot.left), self._height(newRoot.right))
return newRoot
def _rotate_right(self, oldRoot):
newRoot = oldRoot.left #the newRoot is the left child of the previous root
oldRoot.left = newRoot.right #replacing left child of the old root with the right child of the new
if newRoot.right is not None:
newRoot.right.parent = oldRoot
newRoot.parent = oldRoot.parent
if oldRoot == self.root:
self.root = newRoot
else:
if oldRoot.parent.right is oldRoot: #Checking isRightChild
oldRoot.parent.right = newRoot
else:
oldRoot.parent.left = newRoot
newRoot.right = oldRoot
oldRoot.parent = newRoot
oldRoot.height = 1 + max(self._height(oldRoot.left), self._height(oldRoot.right))
newRoot.height = 1 + max(self._height(newRoot.left), self._height(newRoot.right))
return newRoot
if __name__ == '__main__':
obj = BST()
obj.insert('Basic-Instinct', 0)
obj.insert('The-Dreamers', 1)
obj.insert('Saint-Laurent', 2)
obj.insert('Pierrot-le-Fou', 3)
obj.insert('Contempt', 4)
obj.insert('Before-Sunrise', 5)
obj.insert('Cold-War', 8)
obj.display_keys(obj.root) #displays a visual representation of our tree, albeit rotated by 90 degrees
print()
print("isBSTBalanced:", obj.isBSTBalanced())
obj.BalanceTree()
print("isBSTBalanced:", obj.isBSTBalanced()) #After executing BalanceTree(), isBSTBalanced still returns False
print()
obj.display_keys(obj.root)
Progress: Revamped _isBSTBalanced method so that it visits every node recursively and not just the root node. The final outcome, however, remains the same.
Progress: I was able to identify one of the major issues being that while I was calling _rotate_left and _rotate_right methods in the _rebalance method, I was not returning them. In addition to this, I was not recursively visiting the left and right subtrees of curr_node, which was initially set to the root of the tree, to be able to traverse the tree in a bottom up manner. I have resolved this too. I have added a display_keys method which allows us to visualize the tree, albeit rotated by 90 degrees. I'm updating the code and prompt in this post accordingly. The problem that still remains is that I have to call the BalanceTree() method more than once in some cases for isBSTBalanced to return True.

Binary Search Tree Inorder Traversal Python giving recursion Error

I am trying to implement Binary Tree, I think the tree implementation is working fine as I am able to find my elements using if 8 in tree. But I when I am trying to perform inOrder Traversal I am running into Recursion Error.
class BinaryNode:
def init(self,value):
self.value = value
self.left = None
self.right = None
class BinaryTree:
def __init__(self):
self.root = None
def add(self, value):
if self.root == None:
self.root = BinaryNode(value)
else:
current = self.root
while 1:
if value <= current.value:
if current.left:
current = current.left
else:
current.left = BinaryNode(value)
break
elif value > current.value:
if current.right:
current = current.right
else:
current.right = BinaryNode(value)
break
else:
break
def __contains__(self, target):
node = self.root
while node is not None:
if node is not None:
if target < node.value:
node = node.left
elif target > node.value:
node = node.right
else:
return True
return False
def inorder(self,node):
node = self.root
if node is not None:
self.inorder(node.left)
print (node.value)
self.inorder(node.right)
</code>
tree = BinaryTree()
arr = [8,3,1,6]
for i in arr:
tree.add(i)
print (tree.root.value)
print ('Inorder Traversal')
tree.inorder(tree.root)
Error I am getting is "RecursionError: maximum recursion depth exceeded" :( I am trying to check that node is not none before the call, not sure where I am going wrong

Python Serializing/Deserializing a binary tree

I'm trying to implement a serializing/deserializing algorithm in python for binary trees.
Here's my code:
class Node:
count = 1
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def insert(self, value):
if self.value > value:
if self.left is None:
self.left = Node(value)
Node.count += 1
else:
self.left.insert(value)
else:
if self.right is None:
self.right = Node(value)
Node.count += 1
else:
self.right.insert(value)
# Using preorder
def serialize(root, serial):
if root != None:
serial.append(root.value)
serialize(root.left, serial)
serialize(root.right, serial)
else:
serial.append('x')
def deserialize(newRoot, serial):
if serial[0] == 'x':
serial.pop(0)
else:
if len(serial) > 0:
newRoot = Node(serial.pop(0))
print(newRoot.value)
deserialize(newRoot.left, serial)
deserialize(newRoot.right, serial)
print("This program serializes a tree\n")
root = Node(3)
root.insert(1)
root.insert(2)
root.insert(4)
root.insert(5)
root.insert(0)
# Serialize
serial = []
serialize(root, serial)
print(serial)
# Deserialize
newRoot = Node(None)
deserialize(newRoot, serial)
print(newRoot.value)
The problem is, newRoot doesn't get updated by deserialize because python passes it by value. How do I get around this, preferably in the most elegant way? In C/C++, I would just pass a pointer to newRoot and it should get updated accordingly. Thanks!

You can return the newly created nodes and assign them as left and right nodes. Also poping the first element of a list is more costly than poping the last element, so reverseing the list at the beginning and then using it in the recursion will be more performant in your case. So the code will become something like:
def deserialize(serial):
serial.reverse()
return _deserialize(serial)
def _deserialize(serial):
if not serial:
return None
node = None
value = serial.pop()
if value != 'x':
node = Node(value)
node.left = _deserialize(serial)
node.right = _deserialize(serial)
return node
root = deserialize(serial)
print(root.value)

You can create left and right subtree within deserialize function and return the root.
Here is my code:
node_list = []
MARKER = -1
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
def serialize(root):
if root is None:
node_list.append(MARKER)
return
node_list.append(root.val)
serialize(root.left)
serialize(root.right)
def deserialize(root, node_list):
if node_list:
val = node_list.pop(0)
else:
return
if val == MARKER:
return
# Create root, left and right recursively
root = Node(val)
root.left = deserialize(root.left, node_list)
root.right = deserialize(root.right, node_list)
return root
def inorder_traversal(root):
if root:
inorder_traversal(root.left)
print(root.val, end=' ')
inorder_traversal(root.right)
if __name__=="__main__":
# Create tree
root = Node(20)
root.left = Node(8)
root.right = Node(22)
root.left.left = Node(4)
root.left.right = Node(12)
root.left.right.left = Node(10)
root.left.right.right = Node(14)
print("Inorder traversal before serialization..")
inorder_traversal(root)
print('')
# serialize the tree and insert elements into a list
serialize(root)
print(node_list)
root1 = None
root1 = deserialize(root1, node_list)
print("Inorder traversal after deserialization..")
inorder_traversal(root1)
print('')

How can I make a tree Node object display its contents when printed?

This error is occurring when we call the lookup method. Can anyone say how can it be rectified? I am unable to debug it using the available documents online. This is an implementation of a binary tree class. I know it is something related to the equivalence problem.
import deque
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.data = data
def insert(self, data):
if data < self.data:
if self.left is None:
self.left = Node(data)
else:
self.left.insert(data)
else:
if self.right is None:
self.right = Node(data)
else:
self.right.insert(data)
def lookup(self, data, parent=None):
if self.data == data:
return self, parent
if data < self.data:
if self.left is None:
return None
else:
return self.left.lookup(data, parent=self)
else:
if self.right is None:
return None
else:
return self.right.lookup(data, parent=self)
# aka bfs traversal
def level_traversal(self):
root = self
dq = deque()
dq.append(root)
while dq:
root = dq.popleft()
if root.left:
dq.append(root.left)
if root.right:
dq.append(root.right)
print (root.data)
def delete(self, data):
node, parent = self.lookup(data)
if node.children_count() == 0:
if parent.left == node:
parent.left = None
else:
parent.right = None
del node
elif node.children_count() == 1:
if node.left:
n = node.left
else:
n = node.right
if parent:
if parent.left == node:
parent.left = n
else:
parent.right = n
del node
else:
# find the successor
parent = node
successor = node.right
while successor.left:
parent = successor
successor = successor.left
node.data = successor.data
if parent.left == successor:
parent.left = successor.right
else:
parent.right = successor.right
def inorder(self):
if self.left:
self.left.inorder()
print (self.data)
if self.right:
self.right.inorder()
def preorder(self):
print (self.data)
if self.left:
self.left.preorder()
if self.right:
self.right.preorder()
def postorder(self):
if self.left:
self.left.postorder()
if self.right:
self.right.postorder()
print (self.data)
root = Node(8)
root.insert(3)
root.insert(10)
root.insert(1)
root.insert(6)
root.insert(4)
root.insert(7)
root.insert(14)
root.insert(13)
# look up
print (root.lookup(6))
# level traversal
root.level_traversal()
#mirror image
#root.mirror_image()
#root.delete(3)
#root.level_traversal()
# inorder
#root.inorder()
# pre order
#root.preorder()
# postorder
#root.postorder()
# size
#root.size()
#root.dfs()
#print root.height()

This is not an error at all. This happens because you are returning a tuple of objects from your lookup method, and this is just how objects are represented when you print them out. If you don't like this, you can overwrite the __repr__() method.

Try this in your class definition.
Definition for a binary Tree Node
class TreeNode:
def __init__(self, val = 0, left = None, right = None):
self.val = val
self.left = left
self.right = right
## print TreeNode Object
def __repr__(self) -> str:
return '[%s, %r, %r]' % (self.val, self.left, self.right)

Mirror Binary Search Tree

This is a code that given a root of the binary search tree, is to create its mirror.
def mirror(root):
if root is None:
pass
else:
mirror(root.left)
mirror(root.right)
temp = root.left
root.left = root.right
root.right = temp
Firstly, is this code right and also is the recursion here supposed to first get to the leaves of the tree first and then switch the references when unwinding?

It's right, but not very Pythonic.
Better to just write
def mirror(root):
if root is None:
return
mirror(root.left)
mirror(root.right)
root.left, root.right = root.right, root.left
For this problem, you could recurse in either order (either reversing the leaves before the parents or after).

Here is my python code to get mirror image of Binary search tree.There might be some incorrect indentations.
#Binary Tree Node
class Node:
def __init__(self,item):
self.data = item
self.left = None
self.right = None
#Binary search tree
class binarySearchTree:
def __init__(self):
self.root = Node(None)
def mirror(self,node):
if node is None:
return False
elif self.root.data == node:
self.root.left,self.root.right = self.root.right,self.root.left
if self.root.left != None:
self._mirror(self.root.left)
if self.root.right != None:
self._mirror(self.root.right)
else:
return self.root
def _mirror(self,node):
if node is None:
return False
else:
node.left,node.right = node.right,node.left
if node.left != None:
self._mirror(node.left)
if node.right != None:
self._mirror(node.right)
else:
return node
def inorder_traverse(self):
if self.root != None:
self._inorder(self.root)
def _inorder(self,cur):
if cur != None:
if cur.left is not None:
self._inorder(cur.left)
print(cur.data)
if cur.right != None:
self._inorder(cur.right)
def main():
bst = binarySearchTree()
bst.insert(7)
bst.insert(1)
bst.insert(0)
bst.insert(3)
bst.insert(2)
bst.insert(5)
bst.insert(4)
bst.insert(6)
bst.insert(9)
bst.insert(8)
bst.insert(10)
bst.insert(11)
bst.inorder_traverse()
bst.mirror(7)
bst.inorder_traverse()
output:
enter image description here