Python Serializing/Deserializing a binary tree - python

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('')

Related

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.

BST with connection to parent - loop

I have problem with infinite loop in getMinimal() method. It works in this way :
1)Take node,
2)If node has other node on the left - go to other one.
3)Repeat as far as node has sth on the left side
4)Return the minimal node.
But sometimes it works in infinite loop for example from 1000 to 400, then to 4 then..to 1000! I have no ide where I make mistake. I reviewed this code many times,every single "pointer" to parent/left/right node is okay! Please - help.
Algorithm works okay to "handwritten" trees - ~20nodes. I wanted to test it in better cases - 2500nodes,generated by random lib (from -10k to 10k).
import random
class Node:
def __init__(self, val):
self.val = val
self.parent = None
self.right = None
self.left = None
# Class of node.
def str(self):
return str(self.val)
class MyTree:
def __init__(self, node):
self.root = node
def insert(self, node):
current = self.root
a = True
while a:
if node.val > current.val:
if current.right is not None:
current = current.right
continue
else:
current.right = node
node.parent = current
a = False
if node.val <= current.val:
if current.left is not None:
current = current.left
continue
else:
current.left = node
node.parent = current
a = False
def search(self, node):
current = self.root
while node.val != current.val:
if node.val > current.val:
current = current.right
continue
elif node.val <= current.val:
current = current.left
continue
if node.val == current.val:
return current
else:
print("There is no such node!")
def delete(self, node):
if isinstance(node, (float, int)):
node = self.search(node)
if node is self.root:
self.__deleteRoot()
return
else:
if node.right is None and node.left is None:
self.__deleteNN(node)
return
if node.right is None and node.left is not None:
self.__deleteLN(node)
return
if node.right is not None and node.left is None:
self.__deleteNR(node)
return
if node.right is not None and node.left is not None:
self.__deleteLR(node)
return
def __deleteNN(self, node):
if node.parent.left is node:
node.parent.left = None
if node.parent.right is node:
node.parent.right = None
def __deleteLN(self, node):
parent = node.parent
son = node.left
# parent replaced
if parent.left is node:
parent.left = son
if parent.right is node:
parent.right = son
son.parent = parent
def __deleteNR(self,node):
parent = node.parent
son = node.right
# replace parent
if parent.left is node:
parent.left = son
if parent.right is node:
parent.right = son
son.parent = parent
def __deleteLR(self, node):
minimal = self.getMinimal(node.right)
if minimal.parent.left is minimal:
minimal.parent.left = None
if minimal.parent.right is minimal:
minimal.parent.right = None
# parent of minimal done..
if node.parent.left is node:
node.parent.left = minimal
if node.parent.right is node:
node.parent.right = minimal
minimal.right = node.right
minimal.left = node.left
def getMinimal(self, node):
k = node
while k.left is not None:
k = k.left
return k
def getMax(self):
current = self.root
while current.right:
current = current.right
return current
def __trav(self, node):
if not node:
return
print(node.val)
self.__trav(node.left)
self.__trav(node.right)
def printTrav(self):
self.__trav(self.root)
def __deleteRoot(self):
if self.root.left is None and self.root.right is None:
self.root = None
return
if self.root.left is None and self.root.right is not None:
# left empty,right full
self.root.right.parent = None
self.root = self.root.right
return
if self.root.left is not None and self.root.right is None:
# right empty, left full
self.root.left.parent = None
self.root = self.root.left
return
# node has both children
if self.root.left is not None and self.root.right is not None:
temp = self.getMinimal(self.root.right) # minimal from right subtree
# sometimes it could be like this..
# r
# \
# x
if temp.parent.left is temp:
temp.parent.left = None
else:
temp.parent.right = None
self.root.left.parent = temp
self.root.right.parent = temp
temp.right = self.root.right
temp.left = self.root.left
self.root = temp
self.root.parent = None
return
def search(self, val):
node = self.root
if node.val == val:
return node
if val > node.val and node.right is not None:
node = node.right
if val < node.val and node.left is not None:
node = node.left
else:
print("There's no such value!")
return
def printMax(self):
print(self.getMax().val)
def printMin(self):
print(self.getMinimal(self.root).val)
arr=[None]*2500
for x in range(2500):
arr[x]=Node(random.randint(-10000,10000))
myTree = MyTree(arr[0])
for x in range(1,2500):
myTree.insert(arr[x])
for x in range(2500):
myTree.delete(arr[x])
It is suspicious that you define search twice.
Still that said, here is how I would debug this. I would modify your program to read from a file, try to run, and then detect an endless loop and bail out. Now write random files until you have one that causes you to crash.
Once you have a random file that shows the bug, the next step is to make it minimal. Here is a harness that can let you do that.
import itertools
flatten = itertools.chain.from_iterable
# bug_found should be a function that takes a list of elements and runs your test.
# example should be an array that demonstrates the bug.
def find_minimal (bug_found, example):
parts = [example]
while 1 < max(len(part) for part in parts):
i = 0
while i < len(parts):
if 1 == len(parts[i]):
i = i + 1
else:
part = parts.pop(i)
# Divide in 2.
mid = len(part)/2
part1 = part[0:mid]
part2 = part[mid:len(part)]
# Do we need part1?
parts.insert(i, part1)
if bug_found(flatten(parts)):
i = i + 1
parts.insert(i, part2)
else:
parts[i] = part2
# Do we need part2?
if bug_found_func(flatten(parts)):
i = i + 1
else:
parts.pop(i)
return list(flatten(parts))
Just let it run, and after some time it is likely to find a small example. Which will greatly aid in debugging.
So - I found 2 serious bugs in code. Both in LR ("standard" node and root). As I suspected - bugs were in pointers. Now tree is working (tested few times for 20k,30k and 100k nodes). Solved.

Return a node's parent that is exactly below the root

I have a general tree that might look like this:
I want to write a function MyFunc(tree,tree_element)
that will return the element's parent. But not the immediate parent, but a parent that is exactly one level below the root.
In case of the attached tree:
MyFunc(tree,'Dasani')
will return 'Coke', while:
MyFunc(tree,'Zero Sugar')
will return 'Pepsi'
I was able to write two functions.
First returns the immediate parent:
class tree:
def __init__(self, key):
self.data = key
self.left = None
self.right = None
def parent_search(self, root, child_node):
if root :
if root.left and root.left.data== child_node:
return root.data
if root.right and root.right.data== child_node:
return root.data
elif root:
return self.parent_search(root.left, child_node) or self.parent_search(root.right, child_node)
root = tree('Beverage')
root.left = tree('Pepsi')
root.right = tree('Coke')
root.left.left = tree('Zero Sugar')
root.left.right = tree('Cherry Pepsi')
root.right.left = tree('Sport Drinks')
root.right.left.left = tree('Powerade')
root.right.left.right = tree('Dasani')
root.right.left.left.left = tree('Powerade w/Sugar')
print(root.parent_search(root,'Dasani'))
The second returns the whole path all the way up to the root:
def printAncestors(root, target):
if root == None:
return False
if root.data == target:
return True
if (printAncestors(root.left, target) or
printAncestors(root.right, target)):
print root.data,
return True
return False
printAncestors(root, 'Dasani')
However, I need something in between that will return only one element that is exactly one level below the root.
I can calculate the height of the tree from a leaf in question to the root. Then I was thinking to return an element that is height-1 above the leaf but am not sure that it's the right approach.
One way to do that is to pass the path to root to each level of recursion, and then return the entire path, then you can extract whatever part of the path you need, like:
Code:
def parent_search(self, child_node, path_so_far=()):
path_so_far += self.data,
if (self.left and self.left.data == child_node or
self.right and self.right.data == child_node):
return path_so_far
return (
self.left and self.left.parent_search(child_node, path_so_far) or
self.right and self.right.parent_search(child_node, path_so_far)
)
Test Code:
class Tree:
def __init__(self, key):
self.data = key
self.left = None
self.right = None
def parent_search(self, child_node, path_so_far=()):
path_so_far += self.data,
if (self.left and self.left.data == child_node or
self.right and self.right.data == child_node):
return path_so_far
return (
self.left and self.left.parent_search(child_node, path_so_far) or
self.right and self.right.parent_search(child_node, path_so_far)
)
root = Tree('Beverage')
root.left = Tree('Pepsi')
root.right = Tree('Coke')
root.left.left = Tree('Zero Sugar')
root.left.right = Tree('Cherry Pepsi')
root.right.left = Tree('Sport Drinks')
root.right.left.left = Tree('Powerade')
root.right.left.right = Tree('Dasani')
root.right.left.left.left = Tree('Powerade w/Sugar')
print(root.parent_search('Dasani'))
print(root.parent_search('Dasani')[1])
Results:
('Beverage', 'Coke', 'Sport Drinks')
Coke

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

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

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