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148 changes: 123 additions & 25 deletions binary_search_tree/tree.py
Original file line number Diff line number Diff line change
@@ -1,3 +1,7 @@
from multiprocessing.dummy import current_process
from typing import ItemsView

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Look like these import statements never got used. I'm guessing they got added automatically by autocomplete, but you can delete them regardless.



class TreeNode:
def __init__(self, key, val = None):
if val == None:
Expand All @@ -9,50 +13,144 @@ def __init__(self, key, val = None):
self.right = None



class Tree:
def __init__(self):
self.root = None

# Time Complexity:
# Space Complexity:
# Time Complexity: 0(log n)
# Space Complexity: 0(log n)
def add_helper(self, current_node, key, value):
# helper takes advantage of recursive
# if node is empty, return new node with key and value
if current_node == None:
return TreeNode(key, value)
if key <= current_node.key:
current_node.left = self.add_helper(current_node.left, key, value)
else:
current_node.right = self.add_helper(current_node.right, key, value)
return current_node

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This implementation of add_helper works, but definitely does a bit of extra work in setting nodes. Essentially, every node on the path to where the new node ends up will have either its left or right node "updated", though most of these will be updated to the exact same node. The only one that gets a meaningful update is whenever the bottom of the tree is reached and gets its left/right node updated to the new node. These updates happen in lines 28 and 30.

This probably will not be too bad of a performance hit, but there is a way that you can do this and just update only the node you need.

But this overall fine enough!


def add(self, key, value = None):
pass
if self.root == None:
self.root = TreeNode(key, value)
else:
self.add_helper(self.root, key, value)


# Time Complexity:
# Space Complexity:
# Time Complexity: 0(log n)
# Space Complexity: 0(log n)

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Luckily since you did an iterative approach for this, we do not take the hit of the space that will be added to the stack if you had done this recursively. As such, the space complexity is O(1).

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Ignore my comment on how there is space complexity savings for a recursive solution. Both should be O(1) in my opinion because they are not using additional space. I was going off the instructor solution and then realized that I think they made a mistake.

def find(self, key):
pass
if self.root == None:
return None
current = self.root
while current != None:
if current.key == key:
return current.value
elif current.key <key:
current = current.right
else:
current = current.left
return None


# Time Complexity:
# Space Complexity:
# Time Complexity: 0(n)
# Space Complexity: 0(n)
def inorder_helper(self, current_node, items):
if current_node != None:
self.inorder_helper(current_node.left, items)
items.append({"key": current_node.key, "value": current_node.value})
self.inorder_helper(current_node.right, items)

def inorder(self):
pass
#list of items to be returned --> depth-first traversal sorted key values in an ascending order, Left,Current,Right
items = []
self.inorder_helper(self.root, items)
return items


# Time Complexity:
# Space Complexity:
# Time Complexity: 0(n)
# Space Complexity: 0(n)
def preorder_helper(self, current_node, items):
if current_node == None:
return items
else:
items.append({"key": current_node.key, "value": current_node.value})
self.preorder_helper(current_node.left, items)
self.preorder_helper(current_node.right, items)
return items

def preorder(self):
pass
#list of items to be returned --> depth-first traversal sorted current, left, right
items = []
if self.root:
self.preorder_helper(self.root, items)
return items


# Time Complexity: 0(n)
# Space Complexity: 0(n)
def postorder_helper(self, current_node, items):
if current_node:
self.postorder_helper(current_node.left, items)
self.postorder_helper(current_node.right, items)
items.append({"key": current_node.key, "value": current_node.value})
return items

def postorder(self):
#list of items to be returned --> depth-first traversal sorted left, right, current
items = []
if self.root:
self.postorder_helper(self.root, items)
return items

# Time Complexity:
# Space Complexity:
def postorder(self):
pass

# Time Complexity:
# Space Complexity:
# Time Complexity: 0(n)
# Space Complexity: 0(n)

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This is true assuming an unbalanced tree, but if we assume a balanced tree, then we only need to recurse down O(log(n)) times along each branch of the tree, so we can use that for the time and space complexity calculations.

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When I gave this comment, I was working off the instructor solution, but I now think this is wrong. The time complexity is O(n) regardless because we have to travel to every node, and O(1) regardless because no additional space is allocated.

Sorry for any confusion.

def height_helper(self, current_node):
# uses stack and recursion for each node in tree
# plus one for the root node / the current node
if current_node != None:
height_left = self.height_helper(current_node.left)
height_right = self.height_helper(current_node.right)
return (max(height_left, height_right) + 1)

else:
return 0

def height(self):
pass
return self.height_helper(self.root)



# # Optional Method
# # Time Complexity:
# # Space Complexity:
def bfs(self):
pass
# # Time Complexity: O(n)
# # Space Complexity: O(n)

def bfs_helper(self, current_node, sub_list):
if current_node.left:
sub_list.append(current_node.left)
if current_node.right:
sub_list.append(current_node.right)



def bfs(self):
# uses queues to create sublists for each level, pops off elements from queue from left to right until sublist is empty to creates list, then adds children of each element to queue as popping off, repeats,
# recursion for each level
items = []
sub_list = []

if not self.root:
return items

items.append({"key": self.root.key, "value": self.root.value})
self.bfs_helper(self.root, sub_list)

while sub_list:
current_node = sub_list.pop(0)
items.append({"key": current_node.key, "value": current_node.value})
self.bfs_helper(current_node, sub_list)
return items


# # Useful for printing
def to_s(self):
Expand Down