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Scissors - Araceli #56

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153 changes: 129 additions & 24 deletions binary_search_tree/tree.py
Original file line number Diff line number Diff line change
@@ -1,10 +1,12 @@


class TreeNode:
def __init__(self, key, val = None):
if val == None:
val = key
def __init__(self, key, value = None):
if value == None:
value = key

self.key = key
self.value = val
self.value = value
self.left = None
self.right = None

Expand All @@ -14,45 +16,148 @@ class Tree:
def __init__(self):
self.root = None

# Time Complexity:
# Space Complexity:
# Time Complexity: O(n)
# Space Complexity: O(1)
def add(self, key, value = None):
Comment on lines +19 to 21

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👍

pass
tree = TreeNode(key, value)

if self.root == None:
self.root = tree
else:
current = self.root

while current != None:
if key == current.key:
return None
if key < current.key:
if current.left == None:
current.left = tree
current = current.left
elif key > current.key:
if current.right == None:
current.right = tree
current = current.right

# Time Complexity:
# Space Complexity:
# Time Complexity: O(n)
# Space Complexity: O(1)
def find(self, key):
Comment on lines +41 to 43

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👍 Time complexity is O(n) if the tree is unbalanced and O(log n) if it is balanced.

pass

if self.root == None:
return None

current = self.root

while current != None:
if current.key == key:
return current.value
elif key < current.key:
current = current.left
else:
current = current.right

return None

# Time Complexity:
# Space Complexity:
# Time Complexity: O(n)
# Space Complexity: O(1)
def inorder(self):
Comment on lines +60 to 62

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👍 However the space complexity is O(n) because you're building a list of all the nodes.

pass
tree_nodes = []

if self.root == None:
return tree_nodes
else:
return self.inorder_helper(self.root, tree_nodes)

def inorder_helper(self, current, tree_nodes):
if current == None:
return None

self.inorder_helper(current.left, tree_nodes)

tree_nodes.append(
{
"key": current.key,
"value": current.value
}
)

# Time Complexity:
# Space Complexity:
self.inorder_helper(current.right, tree_nodes)

return tree_nodes

# Time Complexity: O(n)
# Space Complexity: O(1)
def preorder(self):
Comment on lines +87 to 89

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👍 However the space complexity is O(n) because you're building a list of all the nodes.

pass
tree_nodes = []

if self.root == None:
return tree_nodes
else:
return self.preorder_helper(self.root, tree_nodes)

# Time Complexity:
# Space Complexity:
def preorder_helper(self, current, tree_nodes):
if current == None:
return None

tree_nodes.append(
{
"key": current.key,
"value": current.value
}
)

self.preorder_helper(current.left, tree_nodes)

self.preorder_helper(current.right, tree_nodes)

return tree_nodes

# Time Complexity: O(n)
# Space Complexity: O(1)
def postorder(self):
Comment on lines +114 to 116

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👍 However the space complexity is O(n) because you're building a list of all the nodes.

pass
tree_nodes = []


return self.postorder_helper(self.root, tree_nodes)

def postorder_helper(self, current, tree_nodes):
if current == None:
return tree_nodes

self.postorder_helper(current.left, tree_nodes)

self.postorder_helper(current.right, tree_nodes)

# Time Complexity:
# Space Complexity:
tree_nodes.append(
{
"key": current.key,
"value": current.value
}
)

return tree_nodes

# Time Complexity: O(n)
# Space Complexity: O(1)
def height(self):
Comment on lines +139 to 141

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👍 Because of the recursion the space complexity is O(h) where h is the height of the tree (which will be log n if the tree is balanced and it can approach n if the tree is very unbalanced).

pass

return self.height_helper(self.root)

def height_helper(self, current):
if current == None:
return 0

left_height = self.height_helper(current.left)

right_height = self.height_helper(current.right)

return max(left_height, right_height) + 1

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




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