awwalm/unique_bst.md

## unique_bst.md

      
    Raw
  

              unique_bst.md
            
          
    Constructing Unique Binary Search Trees

Q: Via Quora

Given the set of keys {6, 7, 1, 4, 2}, how many unique binary search trees of height 2 are possible? Draw the trees. Assume that the height of the root-node is zero.

A:

There are 6 unique Binary Search Trees (BST) of height 2 that can be constructed from the key set
$\{ 6, 7, 1, 4, 2 \}$ when all keys are used. Furthermore, there are 36 total BSTs of height 2
that can be constructed from said key set.

Let’s understand how and why.
Designing a Programmatic Approach


An approach to this problem is to first calculate every possible permutation of the keys
which should give you $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ permutations.


Then attempt to construct a BST for every single permutation. However, once the height of the current BST
exceeds a specified height limit, we halt and skip the current iteration.


The steps hitherto generate all BSTs of the specified height based on the permutated sequences

of the original keys. However, different sequences can still produce the same BST. To address this,
we ensure to perform level-oriented traversal i.e. Breadth-First Search (BFS), for every BST and then
filter and verify if the traversal results had been obtained earlier or not.


The resulting collection of BSTs will be unique, consisting of all elements in the original set of keys, with the exact height specified.


We now introduce a tree printing heuristic that illustrates each BST.


Implementation (see unique_bst.py)

This is a factorial time algorithm (proportional to the size of the original keys),
implementing a brute-force technique in constructing unique BSTs based off a permutation of keys.
Furthermore, this particular pattern is known as backtracking - halting a specific iteration
within an algorithm when an unwanted condition is encountered while exploring every other possible conditional case.
Fun Fact: The resulting BSTs from this particular key set are all AVL-compliant!
Output

All BSTs of height 2 = 36
        6       
       / \        
      /   \       
     /     \      
    2       7   
   / \              
  1   4         

        4       
       / \        
      /   \       
     /     \      
    2       7   
   /       /        
  1       6     

        4       
       / \        
      /   \       
     /     \      
    1       6   
     \       \      
      2       7 

        4       
       / \        
      /   \       
     /     \      
    2       6   
   /         \      
  1           7 

        2       
       / \        
      /   \       
     /     \      
    1       6   
           / \      
          4   7 

        4       
       / \        
      /   \       
     /     \      
    1       7   
     \     /        
      2   6     

Number of unique BSTs of height 2 from set of keys <6, 7, 1, 4, 2> = 6

Further Reading & Improvements

Brute-force and factorial time algorithms should be avoided as much as possible in favor for an algorithmically correct and efficient alternative when feasible. For a potentially efficient solution exploiting combinatorial evaluation, consider this StackOverflow post on a similar question.

  
## unique_bst.py
"""
Given the set of keys {6, 7, 1, 4, 2}, how many unique binary search trees of height 2
are possible? Draw the trees. Assume that the height of the root-node is zero.

From Quora: https://qr.ae/psrOXE
"""
import itertools
from typing import Union, List


class BST:

    # Composition Pattern: Nested private Node instance cannot exist independent of BST object
    class _Node:
        __slots__ = "_key", "_value", "_parent", "_left", "_right", "_height"
        def __init__(self, key, value, parent):
            self._key = key
            self._value = value
            self._parent = parent
            self._left = None
            self._right = None
            self._height = 0
        @property
        def key(self):
            return self._key
        @property
        def value(self):
            return self._value
        @property
        def right(self):
            return self._right
        @property
        def left(self):
            return self._left
        @property
        def height(self):
            return self._height
        @property
        def parent(self):
            return self._parent


    # BST methods and properties (deletion not implemented)
    def __init__(self):
        self._root: Union[BST._Node, None] = None
        self._size = 0

    @property
    def root(self):
        return self._root

    @property
    def size(self):
        return self._size

    def empty(self):
        return self._size == 0

    # noinspection PyMethodMayBeStatic
    def _update_height(self, subtree: _Node):   # Required by tree-printing heuristic
        if (not subtree.left) and (not subtree.right):
            subtree._height = 0
        elif subtree.left and not subtree.right:
            subtree._height = 1 + subtree.left.height
        elif subtree.right and not subtree.left:
            subtree._height = 1 + subtree.right.height
        else:                                   # Both subtrees are present
            subtree._height = 1 + max(subtree.left.height, subtree.right.height)

    def insert(self, key, value):
        max_height = 0
        if self.empty():
            self._root = self._Node(key, value, None)
            self._size = 1
        else:
            subtree, max_height = self.search(self._root, key, value, 0)
            if subtree.key == key:
                subtree._value = value          # Key already present in tree, update only value
                return max_height
            elif key < subtree.key:
                subtree._left = self._Node(key, value, subtree)
            else:                               # key > subtree.key
                subtree._right = self._Node(key, value, subtree)
            self._size += 1
            while subtree:
                self._update_height(subtree)
                subtree = subtree.parent
        return max_height

    def search(self, subtree: _Node, key, value, max_height: int):
        if subtree.key == key and subtree.value == value:
            return subtree, max_height          # Successful search
        elif key < subtree.key:
            if subtree.left:
                return self.search(subtree.left, key, value, max_height+1)
        elif key > subtree.key:
            if subtree.right:
                return self.search(subtree.right, key, value, max_height+1)
        return subtree, max_height              # Unsuccessful search

    def _bfs(self, level: List[_Node], traversed: List[List[_Node]]):
        if len(level) > 0:
            next_level, this_level = [], []
            for node in level:
                if node: this_level.append(node.value)
                if node.left: next_level.append(node.left)
                if node.right: next_level.append(node.right)
            traversed.append(this_level)
            del level                           # Free memory
            self._bfs(next_level, traversed)

    def traverse(self):
        values = []
        self._bfs([self._root], values)
        return values


def print_pretty_tree(start_node):
    """From GitHub: https://github.com/zinchse/AVLTree"""
    space_symbol = r" "
    spaces_count = 4 * 2 ** start_node.height
    out_string = r""
    initial_spaces_string = space_symbol * spaces_count + "\n"
    if not start_node:
        return "Tree is empty"
    height = 2 ** start_node.height
    level = [start_node]

    while len([i for i in level if (not i is None)]) > 0:
        level_string = initial_spaces_string
        for i in range(len(level)):
            j = int((2 * i + 1) * spaces_count / (2 * len(level)))
            level_string = (level_string[:j]
                    + (str(level[i].value) if level[i] else space_symbol)
                    + level_string[j+1:])
        out_string += level_string

        # create next level
        level_next = []
        for i in level:
            level_next += ([i.left, i.right] if i else [None, None])
        # add connection to the next nodes
        for w in range(height - 1):
            level_string = initial_spaces_string
            for i in range(len(level)):
                if not level[i] is None:
                    shift = spaces_count // (2 * len(level))
                    j = (2 * i + 1) * shift
                    level_string = level_string[:j - w - 1] + (
                        '/' if level[i].left else
                        space_symbol) + level_string[j - w:]
                    level_string = level_string[:j + w + 1] + (
                        '\\' if level[i].right else
                        space_symbol) + level_string[j + w:]
            out_string += level_string
        height = height // 2
        level = level_next

    return out_string


def get_unique_bsts(keys: set, max_height: int):
    permutations = itertools.permutations(keys) # All possible permutations of keys
    unique_bsts = set()                         # Unique BSTs of height max_height based on given keys
    bfs_values = list()                         # BFS sequence of keys to verify structural uniqueness
    non_unique_bsts = set()                     # [Optional] All BSTs of height max_height

    for keys in permutations:
        bst = BST()
        height = 0
        for k in keys:
            height = bst.insert(k, k)
            if height > max_height-1:           # Max height exceeded; halt BST construction
                break
        if height == max_height-1:
            non_unique_bsts.add(bst)
            bfs_val = bst.traverse()
            # print(bfs_val)                    # BFS sequence for all BSTs having a height of max_height
            if bfs_val not in bfs_values:
                unique_bsts.add(bst)            # BSTs filtered for uniqueness
                bfs_values.append(bfs_val)

    print(f"All BSTs of height {max_height} = {len(non_unique_bsts)}")

    for tree in unique_bsts: print(print_pretty_tree(tree.root))

    return len(unique_bsts)


if __name__ == '__main__':
    print(
        f"Number of unique BSTs of height 2 from set of keys <6, 7, 1, 4, 2> = "
        f"{get_unique_bsts(keys={6, 7, 1, 4, 2}, max_height=2)}"
    )
	"""
	Given the set of keys {6, 7, 1, 4, 2}, how many unique binary search trees of height 2
	are possible? Draw the trees. Assume that the height of the root-node is zero.

	From Quora: https://qr.ae/psrOXE
	"""
	import itertools
	from typing import Union, List


	class BST:

	# Composition Pattern: Nested private Node instance cannot exist independent of BST object
	class _Node:
	__slots__ = "_key", "_value", "_parent", "_left", "_right", "_height"
	def __init__(self, key, value, parent):
	self._key = key
	self._value = value
	self._parent = parent
	self._left = None
	self._right = None
	self._height = 0
	@property
	def key(self):
	return self._key
	@property
	def value(self):
	return self._value
	@property
	def right(self):
	return self._right
	@property
	def left(self):
	return self._left
	@property
	def height(self):
	return self._height
	@property
	def parent(self):
	return self._parent


	# BST methods and properties (deletion not implemented)
	def __init__(self):
	self._root: Union[BST._Node, None] = None
	self._size = 0

	@property
	def root(self):
	return self._root

	@property
	def size(self):
	return self._size

	def empty(self):
	return self._size == 0

	# noinspection PyMethodMayBeStatic
	def _update_height(self, subtree: _Node): # Required by tree-printing heuristic
	if (not subtree.left) and (not subtree.right):
	subtree._height = 0
	elif subtree.left and not subtree.right:
	subtree._height = 1 + subtree.left.height
	elif subtree.right and not subtree.left:
	subtree._height = 1 + subtree.right.height
	else: # Both subtrees are present
	subtree._height = 1 + max(subtree.left.height, subtree.right.height)

	def insert(self, key, value):
	max_height = 0
	if self.empty():
	self._root = self._Node(key, value, None)
	self._size = 1
	else:
	subtree, max_height = self.search(self._root, key, value, 0)
	if subtree.key == key:
	subtree._value = value # Key already present in tree, update only value
	return max_height
	elif key < subtree.key:
	subtree._left = self._Node(key, value, subtree)
	else: # key > subtree.key
	subtree._right = self._Node(key, value, subtree)
	self._size += 1
	while subtree:
	self._update_height(subtree)
	subtree = subtree.parent
	return max_height

	def search(self, subtree: _Node, key, value, max_height: int):
	if subtree.key == key and subtree.value == value:
	return subtree, max_height # Successful search
	elif key < subtree.key:
	if subtree.left:
	return self.search(subtree.left, key, value, max_height+1)
	elif key > subtree.key:
	if subtree.right:
	return self.search(subtree.right, key, value, max_height+1)
	return subtree, max_height # Unsuccessful search

	def _bfs(self, level: List[_Node], traversed: List[List[_Node]]):
	if len(level) > 0:
	next_level, this_level = [], []
	for node in level:
	if node: this_level.append(node.value)
	if node.left: next_level.append(node.left)
	if node.right: next_level.append(node.right)
	traversed.append(this_level)
	del level # Free memory
	self._bfs(next_level, traversed)

	def traverse(self):
	values = []
	self._bfs([self._root], values)
	return values


	def print_pretty_tree(start_node):
	"""From GitHub: https://github.com/zinchse/AVLTree"""
	space_symbol = r" "
	spaces_count = 4 * 2 ** start_node.height
	out_string = r""
	initial_spaces_string = space_symbol * spaces_count + "\n"
	if not start_node:
	return "Tree is empty"
	height = 2 ** start_node.height
	level = [start_node]

	while len([i for i in level if (not i is None)]) > 0:
	level_string = initial_spaces_string
	for i in range(len(level)):
	j = int((2 * i + 1) * spaces_count / (2 * len(level)))
	level_string = (level_string[:j]
	+ (str(level[i].value) if level[i] else space_symbol)
	+ level_string[j+1:])
	out_string += level_string

	# create next level
	level_next = []
	for i in level:
	level_next += ([i.left, i.right] if i else [None, None])
	# add connection to the next nodes
	for w in range(height - 1):
	level_string = initial_spaces_string
	for i in range(len(level)):
	if not level[i] is None:
	shift = spaces_count // (2 * len(level))
	j = (2 * i + 1) * shift
	level_string = level_string[:j - w - 1] + (
	'/' if level[i].left else
	space_symbol) + level_string[j - w:]
	level_string = level_string[:j + w + 1] + (
	'\\' if level[i].right else
	space_symbol) + level_string[j + w:]
	out_string += level_string
	height = height // 2
	level = level_next

	return out_string


	def get_unique_bsts(keys: set, max_height: int):
	permutations = itertools.permutations(keys) # All possible permutations of keys
	unique_bsts = set() # Unique BSTs of height max_height based on given keys
	bfs_values = list() # BFS sequence of keys to verify structural uniqueness
	non_unique_bsts = set() # [Optional] All BSTs of height max_height

	for keys in permutations:
	bst = BST()
	height = 0
	for k in keys:
	height = bst.insert(k, k)
	if height > max_height-1: # Max height exceeded; halt BST construction
	break
	if height == max_height-1:
	non_unique_bsts.add(bst)
	bfs_val = bst.traverse()
	# print(bfs_val) # BFS sequence for all BSTs having a height of max_height
	if bfs_val not in bfs_values:
	unique_bsts.add(bst) # BSTs filtered for uniqueness
	bfs_values.append(bfs_val)

	print(f"All BSTs of height {max_height} = {len(non_unique_bsts)}")

	for tree in unique_bsts: print(print_pretty_tree(tree.root))

	return len(unique_bsts)


	if __name__ == '__main__':
	print(
	f"Number of unique BSTs of height 2 from set of keys <6, 7, 1, 4, 2> = "
	f"{get_unique_bsts(keys={6, 7, 1, 4, 2}, max_height=2)}"
	)