def __array_to_bst(self, data: list) -> BinaryTreeNode:
"""
Converts a sorted array into a height-balanced BST
Parameters:
data (list): sorted data array
Returns:
BinaryTreeNode: root node of built BST
"""
# length of data in current recursion
n = len(data)
# no input data -> no tree
if n == 0:
return None
# single input data element -> node with value
elif n == 1:
return BinaryTreeNode(data[0])
# multiple data elements are recursively split by the median
else:
# root node is the median
mid = n // 2
root = BinaryTreeNode(data[mid])
# left and right subtrees are built from left and right partition
root.left = self.__array_to_bst(data[:mid])
root.right = self.__array_to_bst(data[mid + 1:])
return root
def __delete(self, root: BinaryTreeNode, key: int) -> BinaryTreeNode:
"""
Delete value from BST by recursively traversing the tree
Parameters:
root (BinaryTreeNode): root node of subtree
val (int): key to delete
Returns:
BinaryTreeNode: root of node of subtree
"""
# if root does not exist, there is no more to traverse
if root is None:
return root
# if key is smaller than root, the key must be in the left subtree
if key < root.data:
root.left = self.__delete(root.left, key)
# if key is greater than root, the key must be in the right subtree
elif key > root.data:
root.right = self.__delete(root.right, key)
# else the key is equal to the root, and the root must be deleted
else:
# if root has no children it is deleted; if root has only one child it becomes the new root
if root.left is None:
child_node = root.right
root = None
return child_node
elif root.right is None:
child_node = root.left
root = None
return child_node
# if node has two children, the inorder succesor is chosen as new root
# inorder-succesor is the smallest in the right subtree
temp = self.__get_min_value(root.right)
# the value of the new root is copied to the current root
root.key = temp.key
# the inorder succesor is then deleted
root.right = self.__delete(root.right, temp.key)
return root
def __invert_tree(self, root: BinaryTreeNode) -> BinaryTreeNode:
"""
Inverts the tree by recusively inverting the subtrees
Parameters:
root (BinaryTreeNode): root node of BST
Returns:
BinaryTreeNode: the root node of the inverted BST
"""
if root is None:
return None
else:
# recursively invert left subtree
left = self.__invert_tree(root.left)
# recursively invert right subtree
right = self.__invert_tree(root.right)
# invert children
root.left = right
root.right = left
return root
def __insert(self, root: BinaryTreeNode, val: int) -> BinaryTreeNode:
"""
Inserts value into BST by recursively traversing the tree
Parameters:
root (BinaryTreeNode): root node of subtree
val (int): value to insert
Returns:
BinaryTreeNode: root of node of subtree
"""
# if no root exists, the new node is the root
if root is None:
return BinaryTreeNode(val)
# if value to insert is smaller than root, traverse down left subtree
if val < root.data:
root.left = self.__insert(root.left, val)
# otherwise, traverse down right subtree
else:
root.right = self.__insert(root.right, val)
return root
stack = [(root, -math.inf, math.inf)]
while True:
current, lower, upper = stack.pop()
if (current.value <= lower) or (current.value >= upper):
return False
if current.left:
stack.append((current.left, lower, current.value))
if current.right:
stack.append((current.right, current.value, upper))
if not stack:
break
return True
tree = BinaryTreeNode(6)
tree.left = BinaryTreeNode(3)
tree.right = BinaryTreeNode(9)
tree.left.left = BinaryTreeNode(1)
tree.left.right = BinaryTreeNode(7)
tree.right.left = BinaryTreeNode(7)
tree.right.right = BinaryTreeNode(10)
print(valid_bst(tree))
def __delete(self, root: BinaryTreeNode, key: int) -> BinaryTreeNode:
"""
Delete value from BST by recursively traversing the tree.
The tree is balanced by traveling up the tree from a newly inserted node until
an unbalance node (z) is encountered. The subtree of the unbalanced node is then balanced
by doing one of four possible operations that rotate the node and its child nodes.
Selecting the right operation is a matter of finding the height difference between
the two subtrees to determine whether they are balanced, and finding the two tallest children
of z (meaning the children that have highest subtress themselves) to determine whether
they are left-left, left-right, right-right or right-left children.
Parameters:
root (BinaryTreeNode): root node of subtree
val (int): key to delete
Returns:
BinaryTreeNode: root of node of subtree
"""
# if root does not exist, there is no more to traverse
if root is None:
return root
# if key is smaller than root, the key must be in the left subtree
if key < root.data:
root.left = self.__delete(root.left, key)
# if key is greater than root, the key must be in the right subtree
elif key > root.data:
root.right = self.__delete(root.right, key)
# else the key is equal to the root, and the root must be deleted
else:
# if root has no children it is deleted; if root has only one child it becomes the new root
if root.left is None:
child_node = root.right
root = None
return child_node
elif root.right is None:
child_node = root.left
root = None
return child_node
# if node has two children, the inorder succesor is chosen as new root
# inorder-succesor is the smallest in the right subtree
temp = self.__get_min_value(root.right)
# the value of the new root is copied to the current root
root.data = temp.data
# the inorder succesor is then deleted
root.right = self.__delete(root.right, temp.data)
# update height attribute of parent node
root.height = 1 + max(self.__get_height(root.left),
self.__get_height(root.right))
# get 'balance factor' (height difference between left and right subtrees)
balance = self.get_height_diff(root)
# balance subtree if unbalanced
# Case 1: left-left rotation
# z is the unbalance node, y is the taller child of z, x is the taller child of y
# - z -
# | |
# - y - T4
# | |
# - x - T3
# | |
# T1 T2
# balance factor is greater than 1, while x and y are both left children
if balance > 1 and self.get_height_diff(root.left) >= 0:
return self.right_rotate(root)
# Case 2: left-right rotation
# z is the unbalance node, y is the taller child of z, x is the taller child of y
# - z -
# | |
# - y - T4
# | |
# T1 - x -
# | |
# T1 T2
# balance factor is greater than 1, while y is a left child and x is a right child
if balance > 1 and self.get_height_diff(root.left) < 0:
root.left = self.left_rotate(root.left)
return self.right_rotate(root)
# Case 3: right-right rotation
# z is the unbalance node, y is the taller child of z, x is the taller child of y
# - z -
# | |
# T1 - y -
# | |
# T2 - x -
# | |
# T3 T4
# balance factor is smaller than 1, while x and y are both right children
if balance < -1 and self.get_height_diff(root.right) <= 0:
return self.left_rotate(root)
# Case 4: right-left rotation
# z is the unbalance node, y is the taller child of z, x is the taller child of y
# - z -
# | |
# T1 - y -
# | |
# - x - T4
# | |
# T2 T3
# balance factor is smaller than 1, while y is a right child and x is a left child
if balance < -1 and self.get_height_diff(root.right) > 0:
root.right = self.right_rotate(root.right)
return self.left_rotate(root)
return root
def __insert(self, root: BinaryTreeNode, data: int):
"""
Inserts a node into the AVL tree while ensuring the tree remains balanced.
The tree is balanced by traveling up the tree from a newly inserted node until
an unbalance node (z) is encountered. The subtree of the unbalanced node is then balanced
by doing one of four possible operations that rotate the node and its child nodes.
Selecting the right operation is a matter of finding the height difference between
the two subtrees to determine whether they are balanced, and comparing the values of
the last two children (x and y) along the traveled path to determine whether they are
left-left, left-right, right-right or right-left children.
Parameters:
root (BinaryTreeNode): root node of subtree
data (int): data value of node to be inserted
"""
# insert new node into tree as in a traditional BST
if not root:
return BinaryTreeNode(data)
elif data < root.data:
root.left = self.__insert(root.left, data)
else:
root.right = self.__insert(root.right, data)
# update height attribute of parent node
root.height = 1 + max(self.__get_height(root.left),
self.__get_height(root.right))
# get 'balance factor' (height difference between left and right subtrees)
balance = self.get_height_diff(root)
# balance subtree if unbalanced
# Case 1: left-left rotation
# z is the unbalance node, y is the first child, and x is the grandchild
# - z -
# | |
# - y - T4
# | |
# - x - T3
# | |
# T1 T2
# balance factor is greater than 1, while x and y are both left children
if balance > 1 and data < root.left.data:
return self.right_rotate(root)
# Case 2: left-right rotation
# z is the unbalance node, y is the first child, and x is the grandchild
# - z -
# | |
# - y - T4
# | |
# T1 - x -
# | |
# T1 T2
# balance factor is greater than 1, while y is a left child and x is a right child
if balance > 1 and data > root.left.data:
root.left = self.left_rotate(root.left)
return self.right_rotate(root)
# Case 3: right-right rotation
# z is the unbalance node, y is the first child, and x is the grandchild
# - z -
# | |
# T1 - y -
# | |
# T2 - x -
# | |
# T3 T4
# balance factor is greater than 1, while x and y are both right children
if balance < -1 and data > root.right.data:
return self.left_rotate(root)
# Case 4: right-left rotation
# z is the unbalance node, y is the first child, and x is the grandchild
# - z -
# | |
# T1 - y -
# | |
# - x - T4
# | |
# T2 T3
# balance factor is smaller than 1, while y is a right child and x is a left child
if balance < -1 and data < root.right.data:
root.right = self.right_rotate(root.right)
return self.left_rotate(root)
return root