def remove(self, item):
"""Removes and returns item from the tree."""
# Helper function to adjust placement of an item
def liftMaxInLeftSubtreeToTop(top):
# Replace top's datum with the maximum datum in the left subtree
# Pre: top has a left child
# Post: the maximum node in top's left subtree
# has been removed
# Post: top.getRoot() = maximum value in top's left subtree
parent = top
currentNode = top.getLeft()
while not currentNode.getRight().isEmpty():
parent = currentNode
currentNode = currentNode.getRight()
top.setRoot(currentNode.getRoot())
if parent == top:
top.setLeft(currentNode.getLeft())
else:
parent.setRight(currentNode.getLeft())
# Begin main part of the method
if self.isEmpty(): return None
# Attempt to locate the node containing the item
itemRemoved = None
preRoot = BinaryTree(None)
preRoot.setLeft(self._tree)
parent = preRoot
direction = 'L'
currentNode = self._tree
while not currentNode.isEmpty():
if currentNode.getRoot() == item:
itemRemoved = currentNode.getRoot()
break
parent = currentNode
if currentNode.getRoot() > item:
direction = 'L'
currentNode = currentNode.getLeft()
else:
direction = 'R'
currentNode = currentNode.getRight()
# Return None if the item is absent
if itemRemoved == None: return None
# The item is present, so remove its node
# Case 1: The node has a left and a right child
# Replace the node's value with the maximum value in the
# left subtree
# Delete the maximium node in the left subtree
if not currentNode.getLeft().isEmpty() \
and not currentNode.getRight().isEmpty():
liftMaxInLeftSubtreeToTop(currentNode)
else:
# Case 2: The node has no left child
if currentNode.getLeft().isEmpty():
newChild = currentNode.getRight()
# Case 3: The node has no right child
else:
newChild = currentNode.getLeft()
# Case 2 & 3: Tie the parent to the new child
if direction == 'L':
parent.setLeft(newChild)
else:
parent.setRight(newChild)
# All cases: Reset the root (if it hasn't changed no harm done)
# Decrement the collection's size counter
# Return the item
self._tree = preRoot.getLeft()
self._size -= 1
return itemRemoved
print "Preorder:"
lyst = []
d.preorder(lyst)
print lyst
print "Inorder:"
lyst = []
d.inorder(lyst)
print lyst
print "Postorder:"
lyst = []
d.postorder(lyst)
print lyst
print "Levelorder:"
lyst = []
d.levelorder(lyst)
print lyst
print "Iterator:"
for item in d: print item,
print "Testing an exception:"
t = a.getLeft()
print "True:", t.isEmpty()
t.setRoot("An item")
def remove(self, item):
"""Removes and returns item from the tree."""
# Helper function to adjust placement of an item
def liftMaxInLeftSubtreeToTop(top):
# Replace top's datum with the maximum datum in the left subtree
# Pre: top has a left child
# Post: the maximum node in top's left subtree
# has been removed
# Post: top.getRoot() = maximum value in top's left subtree
parent = top
currentNode = top.getLeft()
while not currentNode.getRight().isEmpty():
parent = currentNode
currentNode = currentNode.getRight()
top.setRoot(currentNode.getRoot())
if parent == top:
top.setLeft(currentNode.getLeft())
else:
parent.setRight(currentNode.getLeft())
# Begin main part of the method
if self.isEmpty(): return None
# Attempt to locate the node containing the item
itemRemoved = None
preRoot = BinaryTree(None)
preRoot.setLeft(self._tree)
parent = preRoot
direction = 'L'
currentNode = self._tree
while not currentNode.isEmpty():
if currentNode.getRoot() == item:
itemRemoved = currentNode.getRoot()
break
parent = currentNode
if currentNode.getRoot() > item:
direction = 'L'
currentNode = currentNode.getLeft()
else:
direction = 'R'
currentNode = currentNode.getRight()
# Return None if the item is absent
if itemRemoved == None: return None
# The item is present, so remove its node
# Case 1: The node has a left and a right child
# Replace the node's value with the maximum value in the
# left subtree
# Delete the maximium node in the left subtree
if not currentNode.getLeft().isEmpty() \
and not currentNode.getRight().isEmpty():
liftMaxInLeftSubtreeToTop(currentNode)
else:
# Case 2: The node has no left child
if currentNode.getLeft().isEmpty():
newChild = currentNode.getRight()
# Case 3: The node has no right child
else:
newChild = currentNode.getLeft()
# Case 2 & 3: Tie the parent to the new child
if direction == 'L':
parent.setLeft(newChild)
else:
parent.setRight(newChild)
# All cases: Reset the root (if it hasn't changed no harm done)
# Decrement the collection's size counter
# Return the item
self._tree = preRoot.getLeft()
self._size -= 1
return itemRemoved
print "Frontier:", frontier(d)
print "Preorder:"
lyst = []
d.preorder(lyst)
print lyst
print "Inorder:"
lyst = []
d.inorder(lyst)
print lyst
print "Postorder:"
lyst = []
d.postorder(lyst)
print lyst
print "Levelorder:"
lyst = []
d.levelorder(lyst)
print lyst
print "Iterator:"
for item in d:
print item,
print "Testing an exception:"
t = a.getLeft()
print "True:", t.isEmpty()
t.setRoot("An item")
