I know this question has been asked hundreds of times, but I haven't been able to find a question exactly similar to my implementation of binary search trees. I have implemented a basic binary search tree class in Python; the only function that currently is not complete is the delete function. I understand the algorithm for how it is done, and the three different deletion cases. The one restriction I have is that I don't want to use parent pointers, and I am unsure how to keep track of the previous node so that I can set it appropriately based on deletion.
For example, for the case where the node to be deleted has one child, I don't know how I can keep track of the previous node (parent node) so that I can update its left or right child to the children of the node that was deleted.
I don't want to use parent pointers for each node in this deletion function
def delete(self, value):
if self.root:
self.__delete(value, self.root)
def __delete(self, value, curr_node):
#How do I keep track of previous (parent) node here?
if curr_node:
if value < curr_node.value:
self.__delete(value, curr_node.left)
elif value > curr_node.value:
self.__delete(value, curr_node.right)
elif value == curr_node.value:
#Case 1: Node is leaf (no children)
#Case 2: Node has one child: the parent node has its child updated to the deleted node's children
#Case 3: Node has two children
pass
Full code
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def __str__(self):
return str(self.value)
class BST:
def __init__(self):
self.root = None
def insert(self, value):
if not self.root:
self.root = Node(value)
else:
self.__insert(value, self.root)
def __insert(self, value, curr_node):
if value < curr_node.value:
if not curr_node.left:
curr_node.left = Node(value)
else:
self.__insert(value, curr_node.left)
elif value > curr_node.value:
if not curr_node.right:
curr_node.right = Node(value)
else:
self.__insert(value, curr_node.right)
else:
print("Node already exists!")
def lookup(self, value):
if value == self.root.value:
return self.root
else:
self.__lookup(value, self.root)
def __lookup(self, value, curr_node):
if value < curr_node.value:
if value == curr_node.left.value:
return curr_node.left
else:
self.__lookup(value, curr_node.left)
elif value > curr_node.value:
if value == curr_node.right.value:
return curr_node.right
else:
self.__lookup(value, curr_node.right)
else:
print("Value doesn't exist!")
def height(self):
if not self.root:
return 0
else:
return self.__height(self.root)
def __height(self, curr_node):
if not curr_node:
return 0
return 1 + max(self.__height(curr_node.left), self.__height(curr_node.right))
def min_value(self):
if self.root:
return self.__min_value(self.root)
def __min_value(self, curr_node):
if not curr_node.left:
return curr_node
return self.__min_value(curr_node.left)
def max_value(self):
if self.root:
return self.__max_value(self.root)
def __max_value(self, curr_node):
if not curr_node.right:
return curr_node
return self.__max_value(curr_node.right)
def count_nodes(self):
if not self.root:
return 0
else:
return self.__count_nodes(self.root)
def __count_nodes(self, curr_node):
if not curr_node:
return 0
return 1 + self.__count_nodes(curr_node.left) + self.__count_nodes(curr_node.right)
def inorder_traversal(self):
if self.root:
return self.__inorder_traversal(self.root)
def __inorder_traversal(self, curr_node):
path = []
if curr_node:
path.extend(self.__inorder_traversal(curr_node.left))
path.append(curr_node.value)
path.extend(self.__inorder_traversal(curr_node.right))
return path
def __num_children(self, curr_node):
if not curr_node.left and not curr_node.right:
return 0
elif curr_node.right or curr_node.left:
return 1
elif curr_node.right and curr_node.left:
return 2
def delete(self, value):
if self.root:
self.__delete(value, self.root)
def __delete(self, value, curr_node):
#How do I keep track of previous (parent) node here?
if curr_node:
if value < curr_node.value:
self.__delete(value, curr_node.left)
elif value > curr_node.value:
self.__delete(value, curr_node.right)
elif value == curr_node.value:
#Case 1: Node is leaf (no children)
#Case 2: Node has one child: the parent node has its child updated to the deleted node's children
#Case 3: Node has two children
pass
Related
This is my BST code:
class BinarySearchTree(BinaryTree):
def insert(self, value):
parent = None
x = self.root
while(x):
parent = x
if value < x.data:
x = x.left
else:
x = x.right
if parent is None:
self.root = Node(value)
elif value < parent.data:
parent.left = Node(value)
else:
parent.right = Node(value)
def search(self, value):
return self._search(value, self.root)
def _search(self, value, node):
if node is None:
return node
if node.data == value:
return BinarySearchTree(node)
if value < node.data:
return self._search(value, node.left)
return self._search(value, node.right)
and i want it to open a .txt file that have words like this
apple
grape
banana
lemon
and only show the first 4 levels
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.
I am currently working on implementing a Binary search tree but I have a problem with creating the add_node() function. I tried to make it in a recursive way, but the result was weird.
For example- If I run add_node(1), add_node(5), and add_node(0), my tree has only 1, no 5 and 0.
I would be happy if you tell me what the problem is.
def add_node(self, value: int) -> None:
if self.root == None:
self.root = Node(value)
return
else:
return self.add_recursion(self.root, value)
def add_recursion(self, node: Node, value: int) -> None:
if node == None:
node = Node(value)
return
elif value < node.value:
return self.add_recursion(node.left, value)
else:
return self.add_recursion(node.right, value)
When a None value is passed into a function, it is passed by value, not by reference, since... there is no reference.
elif value < node.value:
return self.add_recursion(node.left, value)
else:
return self.add_recursion(node.right, value)
When node.left or node.right is None, a Node ends up being created but not attached to node.
So what you could do is handle the cases where they are None separately.
def add_recursion(self, node: Node, value: int) -> None:
if node == None:
node = Node(value)
elif value < node.value:
if node.left == None:
node.left = Node(value)
else:
self.add_recursion(node.left, value)
else:
if node.right == None:
node.right = Node(value)
else:
self.add_recursion(node.right, value)
While this is workable, it becomes quite ugly. Look at the answer by Locke to see a better way to structure your code.
Also, an additional tip for readability, avoid using return where they aren't necessary such as at the end of a flow.
The issue is how you handle if node.left or node.right does not exist. Since the arguments are copied for each call, setting the value of node in add_recursion has no effect on the tree.
def foo(val):
print("foo start:", val)
val = 5
print("foo end:", val)
bar = 3
foo(bar) # Value of bar is copied for call
print("after foo:", bar) # Prints bar is still 3
I think you might also be getting confused due to how the root node is handled. Here is some starter code on how can handle the initial call to add_recursive.
class Node:
def __init__(self, value: int):
self.value = value
self.left = None
self.right = None
def add_recursive(self, value: int):
# TODO: recursively add to left or right
# For example, here is how you could recursively make a linked list
if self.right is None:
self.right = Node(value)
else:
self.right.add_recursive(value)
class BinarySearchTree:
def __init__(self):
self.root = None
def add_node(self, value: int):
if self.root is None:
self.root = Node(value)
else:
# Call add_recursive on root instead of with root.
self.root.add_recursive(value)
tree = BinarySearchTree()
tree.add_node(1)
tree.add_node(5)
tree.add_node(0)
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
I build a binary tree with python code, now I could print it in order with testTree.printInorder(testTree.root). I have tried to lookup some node ,and the function findNode doesn't work anymore . print testTree.findNode(testTree.root,20) whatever I put in just return None.
class TreeNode:
def __init__(self, value):
self.left = None;
self.right = None;
self.data = value;
class Tree:
def __init__(self):
self.root = None
def addNode(self,node,value):
if node == None:
self.root = TreeNode(value)
else:
if value < node.data:
if node.left == None:
node.left = TreeNode(value)
else:
self.addNode(node.left,value)
else:
if node.right == None:
node.right = TreeNode(value)
else:
self.addNode(node.right,value)
def printInorder(self,node):
if node != None:
self.printInorder(node.left)
print node.data
self.printInorder(node.right)
def findNode(self,node,value):
if self.root != None:
if value == node.data:
return node.data
elif value < node.data and node.left != None:
self.findNode(node.left,value)
elif value > node.data and node.right != None:
self.findNode(node.right,value)
else:
return None
testTree = Tree()
testTree.addNode(testTree.root, 200)
testTree.addNode(testTree.root, 300)
testTree.addNode(testTree.root, 100)
testTree.addNode(testTree.root, 30)
testTree.addNode(testTree.root, 20)
#testTree.printInorder(testTree.root)
print testTree.findNode(testTree.root,20)
Any function without an explicit return will return None.
You have not returned the recursive calls within findNode. So, here.
if value == node.data:
return node.data
elif value < node.data and node.left != None:
return self.findNode(node.left,value)
elif value > node.data and node.right != None:
return self.findNode(node.right,value)
Now, I can't help but thinking this is a bit noisy. You'll always start adding from the root, yes?
testTree.addNode(testTree.root, 200)
You could rather do this
testTree.addNode(200)
And to do that, you basically implement your methods on the TreeNode class instead. So, for the addNode.
You could also "return up" from the recursion, rather than "pass down" the nodes as parameters.
class TreeNode:
def __init__(self, value):
self.left = None
self.right = None
self.data = value
def addNode(self,value):
if self.data == None: # Ideally, should never end-up here
self.data = value
else:
if value < self.data:
if self.left == None:
self.left = TreeNode(value)
else:
self.left = self.left.addNode(value)
else:
if self.right == None:
self.right = TreeNode(value)
else:
self.right = self.right.addNode(value)
return self # Return back up the recursion
Then, in the Tree class, just delegate the addNode responsibility to the root
class Tree:
def __init__(self):
self.root = None
def addNode(self,value):
if self.root == None:
self.root = TreeNode(value)
else:
self.root = self.root.addNode(value)
When you recurse to children in findNode you need to return the result, otherwise the function will implicitly return None:
def findNode(self,node,value):
if self.root != None:
if value == node.data:
return node.data
elif value < node.data and node.left != None:
return self.findNode(node.left,value) # Added return
elif value > node.data and node.right != None:
return self.findNode(node.right,value) # Added return
else:
return None