class BST_Test_Tree_Delete_Element_With_10_Elements(TestCase):
def setUp(self):
self._bst = BinarySearchTree()
self._bst.insert(key=5, obj=5)
self._bst.insert(key=8, obj=8)
self._bst.insert(key=7, obj=7)
self._bst.insert(key=9, obj=9)
self._bst.insert(key=10, obj=10)
self._bst.insert(key=2, obj=2)
self._bst.insert(key=1, obj=1)
self._bst.insert(key=3, obj=3)
self._bst.insert(key=4, obj=4)
self._bst.insert(key=6, obj=6)
self._bst_node_count = 10
def test_delete_element_of_tree_non_existing_element(self):
self._bst.remove(11)
self.assertEqual(self._bst.node_count, self._bst_node_count,
'Tree node count must be 10')
def test_delete_element_of_tree_with_10_node(self):
elements_to_delete = [10, 1, 7, 3, 5, 8, 2, 6, 9]
for element in elements_to_delete:
self._bst.remove(element)
self.assertEqual(self._bst.find(element), None,
'Element found in BST after deleting it!')
self._bst_node_count = self._bst_node_count - 1
self.assertEqual(self._bst.node_count, self._bst_node_count,
'Tree node count must tally after deletion')
def tearDown(self):
self._bst = None
def remove(self, key):
node_to_remove, parent, grand_parent = self._find_node_with_key_for_splay(
key) # need to know to splay
# remove as in BST
BinarySearchTree.remove(self, key=key)
#splay the parent from which we fell off the tree
if parent != None:
self._splay_node_with_key(parent.key)
class BSTHashBucket(collections.MutableMapping):
'''
A hash bucket is used to hold objects that hash to the same value in a hash table. This is hash
bucket using a binary search tree. This masquerades as a python dict in code where it is used.
Since bst is used as bucket datastructure, searches take O log(n) rather than O n
Note: HASHBUCKET ITERATION YIELDS KEYS. not the key value pairs in the bucket.
'''
def __init__(self):
self._bst = BinarySearchTree()
def __len__(self):
'''
The number of elements in the hash bucket
'''
return self._bst.node_count
def get(self, key, default = None):
'''
Get object associated with a key and on key miss return specified default. This is
there in Python dict and this class masquerades as dict, we implement it.
'''
try:
value = self[key]
return value
except KeyError:
return default
def __getitem__(self, key):
value = self._bst.find(key)
if value == None:
raise KeyError('Key Error: %s ' % repr(key))
return value
def __delitem__(self, key):
if key in self._bst:
self._bst.remove(key)
else:
raise KeyError('Key Error: %s ' % repr(key))
def __setitem__(self, key, obj):
if key in self._bst:
self._bst.replace(key, obj)
else:
self._bst.insert(key, obj)
def __iter__(self):
for key, value in self._bst.inorder_traversal_with_stack():
yield key
class BST_Test_Empty_Tree(TestCase):
def setUp(self):
self._bst = BinarySearchTree()
def test_node_count_of_empty_tree(self):
self.assertEqual(self._bst.node_count, 0,
'Empty tree node count must be 0')
def test_find_key_empty_tree(self):
self.assertEqual(self._bst.find(20), None,
'Empty tree find operation must return None')
def test_delete_element_empty_tree(self):
self._bst.remove(20)
def tearDown(self):
self._bst = None
class BST_Test_Tree_With_1_Element(TestCase):
def setUp(self):
self._bst = BinarySearchTree()
self._bst.insert(key=10, obj=10)
def test_node_count_of_tree_with_1_node(self):
self.assertEqual(self._bst.node_count, 1, 'Tree node count must be 1')
def test_find_key_of_tree_with_1_node(self):
self.assertEqual(self._bst.find(10), 10,
'Find operation failed on tree with 1 node')
def test_delete_element_of_tree_with_1_node(self):
self._bst.remove(10)
self.assertEqual(self._bst.node_count, 0,
'Empty tree node count must be 0')
self.assertEqual(self._bst.find(20), None,
'Empty tree find operation must return None')
def tearDown(self):
self._bst = None
print('Tree inorder traversal (key, value) pairs:')
print(traversed)
if __name__ == '__main__':
bst = BinarySearchTree()
print('Node count is %s' % str(bst.node_count))
print('Adding key value pairs (1, 1), (2, 2), .... (6, 6)')
kvpairs = [(5, 5), (2, 2), (7, 7), (1, 1), (3, 3), (9, 9), (8, 8), (4, 4),
(6, 6)]
for kvpair in kvpairs:
bst.insert(key=kvpair[0], obj=kvpair[1])
print_tree_inorder(bst)
print_tree_preorder(bst)
print_tree_postorder(bst)
#remove
element_to_remove = 9
print('removing element %s' % str(element_to_remove))
bst.remove(key=element_to_remove)
print_tree_inorder_using_stack(bst)
#replace obj for a key
key_to_replace = 1
new_object_for_Key = 111
bst.replace(key=key_to_replace, obj=new_object_for_Key)
print_tree_inorder_using_stack(bst)