def reverse(num):
str_num = str(num)
stack = LinkedStack(list(str_num))
reversed_num = ""
while not stack.is_empty():
reversed_num += stack.pop()
return int(reversed_num)
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) since we visit every node
Memory usage: O(h) where h is the height of the tree"""
# -create an empty stack and push the root node in there
# -loop through while the stack isn't empty
# -pop() on the stack
# -push() the right node of the node we just popped off the stack
# -pop() and left node of the node we just popped off the stack
stack = LinkedStack()
stack.push(
node
) # since we're starting in the middle, we start the stack off with self.root
while not stack.is_empty():
toPop = stack.pop()
if toPop == None:
continue # skip this one
visit(toPop.data) # add in the data
stack.push(
toPop.right
) # add the right before the left since this is a stack,
stack.push(
toPop.left
) # the next time we look in the stack, the left will be seen before the right!!
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: O(n) we visit every node in the tree
TODO: Memory usage: O(n) because we are creating a stack
"""
# stack to use to store nodes in order
stack = LinkedStack()
# set current to the starting node
current = node
# load starting node into stack
stack.push(current)
# traverse until stack is empty
while not stack.is_empty():
# pop and visit top node in stack
current = stack.pop()
visit(current.data)
# NOTE: right is first because we are using stack
# so right child will be under left child
# if right child is present add to stack
if current.right is not None:
stack.push(current.right)
# if left child is present add to stack
if current.left is not None:
stack.push(current.left)
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: ??? Why and under what conditions?
TODO: Memory usage: ??? Why and under what conditions?"""
stack = LinkedStack()
stack.push(node)
while stack.is_empty() is False:
# add to stack while node.left exist
if node.left is not None:
# add left, or smaller, node on top
stack.push(node.left)
# iterate left tree
node = node.left
# begin iterating right tree
else:
"""once the left tree of node is None, we can begin popping and iterating the right side"""
# visit the top node in the stack
node = stack.pop()
visit(node.data)
# iterate the right node since the left node is now empty
if node.right is not None:
stack.push(node.right)
node = node.right
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: o(n) depends on the number of items in the tree
Memory usage: o(h) the longest the stack will get is the height of the lowest node."""
# Traverse post-order without using recursion (stretch challenge)
stack = LinkedStack()
stack.push(node)
while len(stack) != 0:
pass
def test_pop(self):
s = Stack(['A', 'B', 'C'])
assert s.pop() == 'C'
assert s.length() == 2
assert s.pop() == 'B'
assert s.length() == 1
assert s.pop() == 'A'
assert s.length() == 0
assert s.is_empty() is True
with self.assertRaises(ValueError):
s.pop()
def reverse(nums):
#declare your stack
Stack = LinkedStack() #init the stack
#declare a variable to get track of the reversed number
reversed = []
#add each number in nums to the stack
for num in nums: # Loop through the whole list O(n)
if num == '-':
reversed.append(num) #adds sign to list O(1)
if num.isdigit():
Stack.push(num) #adds num to stack O(1)
#traverse through the stack
while Stack.length() > 0: # Loops through the whole stack
#add the head of the stack then delete the head
reversed.append(Stack.peek()) # adds number back to list O(1)
Stack.pop() # deletes current stack to move to the next one O(1)
if reversed[0] == '-':
if reversed[1] == '0':
reversed.pop(1) #removes a digit O(1)
if reversed[0] == '0':
reversed.pop(0) #removes a digit O(1)
return "".join(reversed)
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) we are visiting each node
Memory usage: O(n) creating a stack with number of nodes in the tree"""
stack = LinkedStack()
stack.push(node)
while not stack.is_empty():
node = stack.pop()
visit(node.data)
if (node.right != None):
stack.push(node.right)
if (node.left != None):
stack.push(node.left)
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: ??? Why and under what conditions?
TODO: Memory usage: ??? Why and under what conditions?"""
# Traverse post-order without using recursion (stretch challenge)
stack = LinkedStack()
stack.push(node)
while not stack.is_empty():
node = stack.pop()
visit(node.data)
if node.left:
stack.push(node.left)
if node.right:
stack.push(node.right)
def topoSort(g, startLabel=None):
stack = LinkedStack()
g.clearVertexMarks()
for v in g.vertices():
if not v.isMarked():
dfs(g, v, stack)
lyst = []
return stack
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(3n) because each node is called recursively 3 times.
Reduces to O(n)
Memory usage: If tree is balanced, height of tree is log(n) and memory
usage is O(log(n))
If Tree is unblanced, height of tree is approx n and
memory usasge is O(n)"""
# Traverse pre-order without using recursion (stretch challenge)
stack = LinkedStack()
stack.push(node)
while not stack.is_empty():
node = stack.pop()
visit(node.data)
if node.right:
stack.push(node.right)
if node.left:
stack.push(node.left)
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: Why and under what conditions?
TODO: Memory usage: ??? Why and under what conditions?"""
# Traverse post-order without using recursion
stack = LinkedStack()
# loop until root node is the current node
stack.push(node)
while not stack.is_empty():4
node = stack.pop() # look at the top of the stack
visit(node.data)
# push the right node to the top of the stack if it is not none and it's not in set
if node.left is not None:
stack.push(node.left)
# push the right node to the top of the stack if it is not none and it's not in set
if node.right is not None:
stack.push(node.right)
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: ??? Why and under what conditions?
TODO: Memory usage: ??? Why and under what conditions?"""
# Traverse pre-order without using recursion (stretch challenge)
stack = LinkedStack()
# Enstack given starting node
stack.push(node)
# Loop until queue is empty
while not stack.is_empty():
# Dequeue node at front of queue
node = stack.pop()
# Visit this node's data with given function
visit(node.data)
# Enqueue this node's left child, if it exists
if node.right:
stack.push(node.right)
# Enqueue this node's right child, if it exists
if node.left:
stack.push(node.left)
def _traverse_in_order_iterative(self, node, visit):
"""
Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: O(n) we visit every node in the tree
TODO: Memory usage: O(n) because we are creating a stack
"""
# stack for holding horizon
stack = LinkedStack()
# tells when we are done traversing
done = False
# set node as start node to start
current = node
# traverse nodes while stack is not empty
while not done:
# traverse down left side of tree
# appending nodes to stack
if current is not None:
stack.push(current)
current = current.left
# if node is none time to go grab
# from stack
else:
# if stack is not empty pop and grab
# top item in stack
if not stack.is_empty():
current = stack.pop()
# visit the node
visit(current.data)
# grab the right node (None is fine)
current = current.right
else:
# All done!
done = True
def _traverse_post_order_iterative(self, node):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) we are visiting each node
Memory usage: O(n) creating a stack with number of nodes in the tree"""
'''
# Using set and stack
stack = Stack()
stack.push(node)
visited = set()
while not stack.is_empty():
if(stack.peek().left != None and stack.peek().left.data not in visited):
stack.push(stack.peek().left)
elif(stack.peek().right != None and stack.peek().right.data not in visited):
stack.push(stack.peek().right)
else:
node = stack.pop()
visit(node.data)
visited.add(node.data)'''
# create a list equal to size of tree
items = [None] * self.size
i = self.size # i is the last node
stack = LinkedStack()
# add the root node to stack
stack.push(node)
# traverse through stack is empty
while not stack.is_empty():
i -= 1
node = stack.pop()
# i is the last index and assign it to data at the root
items[i] = node.data
# check if node has left child and add to stack if it is
if (node.left != None):
stack.push(node.left)
# check if node has right child and add to stack if it is
if (node.right != None):
stack.push(node.right)
return items
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) since we visit every node
Memory usage: O(logn) for the depth of the tree"""
# -create an empty stack
# -curr variable which starts off with the self.root (will be updated in a while loop!!)
# -go in a while loop while true (will be false when the stack is empty)
# -call push() on curr while there's a value
# -update the curr to become the curr's left
# -once we hit a None value:
# - we'd want to pop() off the stack
# - append it to the list
# - update curr to be the recently popped node's right node
# - however, if the stack is empty in here, we set the while loop to false and exit the function!
stack = LinkedStack() # will act as the recursive call
curr = self.root # start at the root
stillLoop = True # loop until the stack is empty
while stillLoop:
# we're at the end of a leaf node
if curr == None:
if stack.length() > 0:
node = stack.pop()
visit(node.data)
curr = node.right
# our stack is empty so we can finally return once all the nodes have been visited
else:
stillLoop = False
# we can still go down the left side
else:
stack.push(curr)
curr = curr.left
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) (n is # of nodes) Worst case since we are visiting every node.
Memory usage: O(n) worst case if unbalanced - O(log n) if balanced since we will push n nodes in the recursive stack"""
# Variables and pointers to keep track of nodes and status as we traverse
curr_node = node
stack = LinkedStack()
done = False
while not done: # Traverse until the leaf on the far right
if curr_node is not done:
stack.push(curr_node
) # Keep track of this current node for future use
curr_node = curr_node.left # Go to the left subtree
else:
if not stack.is_empty():
curr_node = stack.pop(
) # grab the parent node that was pushed
visit(curr_node) # visit its data
curr_node = curr_node.right # go to the right tree
else:
done = True # The entire tree has been traversed in order
def _traverse_pre_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative pre-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: O(n) where n is number of nodes in tree.
TODO: Memory usage: at most log(n) items since we're canceling out on the stack each time traversing"""
# TODO: Traverse pre-order without using recursion (stretch challenge)
stack = LinkedStack([node])
while stack.is_empty() == False:
popped = stack.pop()
if popped == None:
continue
visit(popped.data)
stack.push(popped.right)
stack.push(popped.left)
def bracketsBalance(exp):
"""exp represents the expression"""
stk = LinkedStack() # Create a new stack
for ch in exp: # Scan across the expression
if ch in ['[', '(']: # Push an opening bracket
stk.push(ch)
elif ch in [']', ')']: # Process a closing bracket
if stk.isEmpty(): # Not balanced
return False
chFromStack = stk.pop()
# Brackets must be of same type and match up
if ch == ']' and chFromStack != '[' or \
ch == ')' and chFromStack != '(':
return False
return stk.isEmpty() # They all matched up
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: O(n), have to visit every node
TODO: Memory usage: O(n) nodes are stored on stack"""
# TODO: Traverse in-order without using recursion (stretch challenge)
stack = LinkedStack()
node = self.root
while ((node is not None) or (not stack.is_empty())):
if node is not None:
stack.push(node)
node = node.left
elif not stack.is_empty():
node = stack.pop()
visit(node.data)
node = node.right
def bracketsBalance(exp):
"""exp represents the expression"""
stk = LinkedStack() # Create a new stack
for ch in exp: # Scan across the expression
if ch in ['[', '(']: # Push an opening bracket
stk.push(ch)
elif ch in [']', ')']: # Process a closing bracket
if stk.isEmpty(): # Not balanced
return False
chFromStack = stk.pop()
# Brackets must be of same type and match up
if ch == ']' and chFromStack != '[' or \
ch == ')' and chFromStack != '(':
return False
return stk.isEmpty() # They all matched up
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) where n is number of nodes since having to visit each node.
Memory usage: log(n) items since queue has to store only until at most its subtree until it cancels out?"""
stack = LinkedStack()
while True:
if node == None:
if stack.is_empty():
break
popped = stack.pop()
visit(popped.data)
node = popped.right
else:
stack.push(node)
node = node.left
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(3n) because each node is called 3 times.
Reduces to O(n)
Memory usage: If tree is balanced, height of tree is approx log(n) and memory
usage is O(log(n))
If Tree is unblanced, height of tree is approx n and
memory usasge is O(n)"""
# Traverse in-order without using recursion (stretch challenge)
stack = LinkedStack()
while True:
if node == None:
if stack.is_empty():
break
pop = stack.pop()
visit(pop.data)
node = pop.right
else:
stack.push(node)
node = node.left
def test_peek(self):
s = Stack()
assert s.peek() is None
s.push('A')
assert s.peek() == 'A'
s.push('B')
assert s.peek() == 'B'
s.pop()
assert s.peek() == 'A'
s.pop()
assert s.peek() is None
def test_init(self):
s = Stack()
assert s.peek() is None
assert s.length() == 0
assert s.is_empty() is True
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n)
Memory usage: O(n)
Credit to Nicolai and Ryan for solving what I began"""
# Traverse post-order without using recursion (stretch challenge)
traversed = set()
# Create queue to store nodes not yet traversed
stack = LinkedStack()
# Enqueue root node as last
stack.push(node)
while not stack.is_empty():
node = stack.peek()
if node.right and node.right not in traversed:
stack.push(node.right)
if node.left and node.left not in traversed:
stack.push(node.left)
if (
node.is_leaf()
or node.left is None and node.right in traversed
or node.left in traversed and node.right is None
or node.left in traversed and node.right in traversed
):
node = stack.pop()
visit(node.data)
traversed.add(node)
def _traverse_in_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) we are visiting each node
Memory usage: O(n) creating a stack with number of nodes in the tree"""
stack = LinkedStack()
stack.push(node) # push the root node
while not stack.is_empty():
if (stack.peek().left != None): # if node has left child
stack.push(stack.peek().left)
else:
node = stack.pop()
visit(node.data)
if (not stack.is_empty()):
node = stack.pop()
visit(node.data)
if (node.right is not None):
stack.push(node.right)
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) since we visit every node
Memory usage: O(h) where h is the height of the tree"""
# NOT WORKING
# leftStack = LinkedStack()
# rightStack = LinkedStack()
# theRoot = node # stays the same until end
# curr = node # will update this variable in the while loop
# # visit all the left nodes
# leftStack.push(theRoot.left)
# while not leftStack.is_empty():
# toPop = leftStack.pop()
# if toPop == None:
# continue
# visit(toPop.data)
# leftStack.push(toPop.right)
# leftStack.push(toPop.left)
# print(leftStack.list)
# # visit all the right nodes
# rightStack.push(theRoot.right)
# while not rightStack.is_empty():
# toPop = rightStack.pop()
# if toPop == None:
# continue
# visit(toPop.data)
# rightStack.push(toPop.right)
# rightStack.push(toPop.left)
# # before returning, we add the mid node (self.root)
# visit(theRoot.data)
# WORKING
# -create 2 stacks where the second stack will be what we return starting from the peek to the end
# -start the first stack with the root node
# -loop through first stack while isn't empty
# - pop from the first stack then add it to our second stack
# - push left of the popped item
# - push right of the popped item
# -after our second stack has all the nodes, we can append to the list as we call pop() method
startStack = LinkedStack()
endStack = LinkedStack()
startStack.push(node)
while not startStack.is_empty():
toPop = startStack.pop()
if toPop == None:
continue
endStack.push(toPop)
startStack.push(toPop.left)
startStack.push(toPop.right)
while not endStack.is_empty():
visit(endStack.pop().data)
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
Running time: O(n) where n is number of nodes in tree.
Memory usage: at most log(n) items since stack having to store at most until the furthest down path."""
# TODO: Traverse post-order without using recursion (stretch challenge)
stack = LinkedStack([node])
stack2 = LinkedStack()
while stack.is_empty() == False:
popped = stack.pop()
if popped == None:
continue
stack2.push(popped)
stack.push(popped.left)
stack.push(popped.right)
# pop items from stack2 and append to items list
while stack2.is_empty() == False:
visit(stack2.pop().data)
def _traverse_post_order_iterative(self, node, visit):
"""Traverse this binary tree with iterative post-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: O(n) because every node is visited
TODO: Memory usage: O(log n) we are creating a stack that will hold
at most the same number of nodes as height in tree
"""
# stack to keep track of nodes in order
stack = LinkedStack()
# done keeps track of if done traversing
# controls while loop
done = False
# set current node to starting node
current = node
# traverse nodes while stack is not empty
while not done:
# while current node is something
while current:
# if right child is there push onto stack
if current.right is not None:
stack.push(current.right)
# push current node onto stack (after right children)
stack.push(current)
# grab left child of root
current = current.left
# Current node is none pop node from top of stack
current = stack.pop()
# check if the node has a right child that hasn't
# been checkout yet then push onto stack before current node
if current.right is not None and stack.peek() == current.right:
# pop child node from stack
stack.pop()
# push current node onto stack
stack.push(current)
current = current.right # now set current to right child
else:
# visit the node
visit(current.data)
# set current node to None
current = None
if stack.is_empty():
# stack is empty we are done traversing
done = True
def _traverse_in_order_iterative(self):
"""Traverse this binary tree with iterative in-order traversal (DFS).
Start at the given node and visit each node with the given function.
TODO: Running time: ??? Why and under what conditions?
TODO: Memory usage: ??? Why and under what conditions?"""
# TODO: Traverse in-order without using recursion (stretch challenge)
items = []
stack = LinkedStack()
node = self.root
stack.push(node)
while stack.is_empty() == False:
if node.is_leaf():
items.append(node.data)
stack.pop()
if stack.is_empty() == False:
items.append(stack.peek().data)
node = stack.pop().right
stack.push(node)
else:
if node.left is not None:
node = node.left
stack.push(node)
return items
