Binary Search Tree

class BSTNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def insert(self, value):
        # create a node for the new value
        new_node = BSTNode(value)
        # compare the node value to the self value
        if (
            new_node.value <= self.value
        ):  # less then or equal to will all go to the left if any duplicats too
            # we add to the left
            if self.left is None:
                self.left = new_node
            else:
                self.left.insert(value)
        else:
            # we add to the right
            # if space exists
            if self.right is None:
                self.right = new_node
            else:
                self.right.insert(value)

    def search(self, target):
        if target == self.value:
            return True
        # check if value is less than self.value

        if target < self.value:
            # look to the left
            if self.left is None:
                return False
            return self.left.search(target)

        else:
            # look to the right
            if self.right is None:
                return False
            return self.right.search(target)

    def find_minimum_value(self):
        # if self.left is None: #recursive here
        #     return self.value
        # min_value =  self.left.find_minimum_value()
        # return min_value
        curr_node = self
        while curr_node.left is not None:
            curr_node = curr_node.left
        return curr_node.value


root = BSTNode(10)
# root.left = BSTNode(6)
# root.right = BSTNode(12)
root.insert(6)
root.insert(7)
root.insert(12)
root.insert(5)
root.insert(14)
root.insert(8)

print(f"minimum value in tree is: {root.find_minimum_value()}")

print(f"does 8 exist? {root.search(8)}")
print(f"does 8 exist? {root.search(7)}")
print(f"does 8 exist? {root.search(15)}")
class BSTNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def insert(self, value):
        # create a node for the new value
        new_node = BSTNode(value)
        # compare the node value to the self value
        if (
            new_node.value <= self.value
        ):  # less then or equal to will all go to the left if any duplicats too
            # we add to the left
            if self.left is None:
                self.left = new_node
            else:
                self.left.insert(value)
        else:
            # we add to the right
            # if space exists
            if self.right is None:
                self.right = new_node
            else:
                self.right.insert(value)

    def search(self, target):
        if target == self.value:
            return True
        # check if value is less than self.value

        if target < self.value:
            # look to the left
            if self.left is None:
                return False
            return self.left.search(target)

        else:
            # look to the right
            if self.right is None:
                return False
            return self.right.search(target)

    def find_minimum_value(self):
        # if self.left is None: #recursive here
        #     return self.value
        # min_value =  self.left.find_minimum_value()
        # return min_value
        curr_node = self
        while curr_node.left is not None:
            curr_node = curr_node.left
        return curr_node.value


root = BSTNode(10)
# root.left = BSTNode(6)
# root.right = BSTNode(12)
root.insert(6)
root.insert(7)
root.insert(12)
root.insert(5)
root.insert(14)
root.insert(8)

print(f"minimum value in tree is: {root.find_minimum_value()}")

print(f"does 8 exist? {root.search(8)}")
print(f"does 8 exist? {root.search(7)}")
print(f"does 8 exist? {root.search(15)}")class BSTNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def insert(self, value):
        # create a node for the new value
        new_node = BSTNode(value)
        # compare the node value to the self value
        if (
            new_node.value <= self.value
        ):  # less then or equal to will all go to the left if any duplicats too
            # we add to the left
            if self.left is None:
                self.left = new_node
            else:
                self.left.insert(value)
        else:
            # we add to the right
            # if space exists
            if self.right is None:
                self.right = new_node
            else:
                self.right.insert(value)

    def search(self, target):
        if target == self.value:
            return True
        # check if value is less than self.value

        if target < self.value:
            # look to the left
            if self.left is None:
                return False
            return self.left.search(target)

        else:
            # look to the right
            if self.right is None:
                return False
            return self.right.search(target)

    def find_minimum_value(self):
        # if self.left is None: #recursive here
        #     return self.value
        # min_value =  self.left.find_minimum_value()
        # return min_value
        curr_node = self
        while curr_node.left is not None:
            curr_node = curr_node.left
        return curr_node.value


root = BSTNode(10)
# root.left = BSTNode(6)
# root.right = BSTNode(12)
root.insert(6)
root.insert(7)
root.insert(12)
root.insert(5)
root.insert(14)
root.insert(8)

print(f"minimum value in tree is: {root.find_minimum_value()}")

print(f"does 8 exist? {root.search(8)}")
print(f"does 8 exist? {root.search(7)}")
print(f"does 8 exist? {root.search(15)}")
class BSTNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def insert(self, value):
        # create a node for the new value
        new_node = BSTNode(value)
        # compare the node value to the self value
        if (
            new_node.value <= self.value
        ):  # less then or equal to will all go to the left if any duplicats too
            # we add to the left
            if self.left is None:
                self.left = new_node
            else:
                self.left.insert(value)
        else:
            # we add to the right
            # if space exists
            if self.right is None:
                self.right = new_node
            else:
                self.right.insert(value)

    def search(self, target):
        if target == self.value:
            return True
        # check if value is less than self.value

        if target < self.value:
            # look to the left
            if self.left is None:
                return False
            return self.left.search(target)

        else:
            # look to the right
            if self.right is None:
                return False
            return self.right.search(target)

    def find_minimum_value(self):
        # if self.left is None: #recursive here
        #     return self.value
        # min_value =  self.left.find_minimum_value()
        # return min_value
        curr_node = self
        while curr_node.left is not None:
            curr_node = curr_node.left
        return curr_node.value


root = BSTNode(10)
# root.left = BSTNode(6)
# root.right = BSTNode(12)
root.insert(6)
root.insert(7)
root.insert(12)
root.insert(5)
root.insert(14)
root.insert(8)

print(f"minimum value in tree is: {root.find_minimum_value()}")

print(f"does 8 exist? {root.search(8)}")
print(f"does 8 exist? {root.search(7)}")
print(f"does 8 exist? {root.search(15)}")
# Implement a Binary Search Tree (BST) that can insert values and check if
# values are present

class Node(object):
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

class BST(object):
    def __init__(self, root):
        self.root = Node(root)

    def insert(self, new_val):
        if(self.root.left==None):
            if(self.root.value>new_val):
                self.root.left = Node(new_val)
        elif(self.root.right==None):
            if(self.root.value<new_val):
                self.root.right = Node(new_val)
        else:
            current = self.root
            while(current.left!=None or current.right!=None):
                if(current.value>new_val):
                    current = current.left
                else:
                    current = current.right

            if(current.left==None):
                current.left = Node(new_val)
            else:
                current.right = Node(new_val)

    def search(self, find_val):
        if(self.root.left==None and self.root.right==None and self.root.value!=find_val):
            return False
        else:
            current = self.root
            val_possible = True
            while(val_possible):
                if(current.value==find_val):
                        return True
                if(current.value<find_val):
                    current = current.right
                else:
                    current = current.left
                if(current==None):
                    return False
                if(current.value<find_val and (current.right==None or current.right>find_val)):
                    return False
                if(current.value>find_val and (current.left==None or current.left<find_val)):
                    return False

# Set up tree
tree = BST(4)

# Insert elements
tree.insert(2)
tree.insert(1)
tree.insert(3)
tree.insert(5)

# Check search
# Should be True
print tree.search(4)
# Should be False
print tree.search(6)

The following is the definition of Binary Search Tree(BST) according to Wikipedia Binary Search Tree is a node-based binary tree data structure which has the following properties:

  • The left subtree of a node contains only nodes with keys lesser than the node’s key.

  • The right subtree of a node contains only nodes with keys greater than the node’s key.

  • The left and right subtree each must also be a binary search tree. There must be no duplicate nodes.

The above properties of Binary Search Tree provides an ordering among keys so that the operations like search, minimum and maximum can be done fast. If there is no ordering, then we may have to compare every key to search for a given key.

Searching a key For searching a value, if we had a sorted array we could have performed a binary search. Let’s say we want to search a number in the array what we do in binary search is we first define the complete list as our search space, the number can exist only within the search space. Now we compare the number to be searched or the element to be searched with the mid element of the search space or the median and if the record being searched is lesser we go searching in the left half else we go searching in the right half, in case of equality we have found the element. In binary search we start with ‘n’ elements in search space and then if the mid element is not the element that we are looking for, we reduce the search space to ‘n/2’ and we go on reducing the search space till we either find the record that we are looking for or we get to only one element in search space and be done with this whole reduction. Search operation in binary search tree will be very similar. Let’s say we want to search for the number, what we’ll do is we’ll start at the root, and then we will compare the value to be searched with the value of the root if it’s equal we are done with the search if it’s lesser we know that we need to go to the left subtree because in a binary search tree all the elements in the left subtree are lesser and all the elements in the right subtree are greater. Searching an element in the binary search tree is basically this traversal in which at each step we will go either towards left or right and hence in at each step we discard one of the sub-trees. If the tree is balanced, we call a tree balanced if for all nodes the difference between the heights of left and right subtrees is not greater than one, we will start with a search space of **‘n’**nodes and when we will discard one of the sub-trees we will discard ‘n/2’ nodes so our search space will be reduced to ‘n/2’ and then in the next step we will reduce the search space to ‘n/4’ and we will go on reducing like this till we find the element or till our search space is reduced to only one node. The search here is also a binary search and that’s why the name binary search tree.

# A utility function to search a given key in BST
def search(root,key):
     
    # Base Cases: root is null or key is present at root
    if root is None or root.val == key:
        return root
 
    # Key is greater than root's key
    if root.val < key:
        return search(root.right,key)
   
    # Key is smaller than root's key
    return search(root.left,key)

Illustration to search 6 in below tree:

  1. Start from the root.

  2. Compare the searching element with root, if less than root, then recurse for left, else recurse for right.

  3. If the element to search is found anywhere, return true, else return false.

Insertion of a key A new key is always inserted at the leaf. We start searching a key from the root until we hit a leaf node. Once a leaf node is found, the new node is added as a child of the leaf node.

         100                               100
        /   \        Insert 40            /    \
      20     500    --------->          20     500 
     /  \                              /  \  
    10   30                           10   30
                                              \   
                                              40
# Python program to demonstrate
# insert operation in binary search tree
 
# A utility class that represents
# an individual node in a BST
 
 
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
# A utility function to insert
# a new node with the given key
 
 
def insert(root, key):
    if root is None:
        return Node(key)
    else:
        if root.val == key:
            return root
        elif root.val < key:
            root.right = insert(root.right, key)
        else:
            root.left = insert(root.left, key)
    return root
 
# A utility function to do inorder tree traversal
 
 
def inorder(root):
    if root:
        inorder(root.left)
        print(root.val)
        inorder(root.right)
 
 
# Driver program to test the above functions
# Let us create the following BST
#    50
#  /     \
# 30     70
#  / \ / \
# 20 40 60 80
 
r = Node(50)
r = insert(r, 30)
r = insert(r, 20)
r = insert(r, 40)
r = insert(r, 70)
r = insert(r, 60)
r = insert(r, 80)
 
# Print inoder traversal of the BST
inorder(r)

Output

20
30
40
50
60
70
80

Illustration to insert 2 in below tree:

  1. Start from the root.

  2. Compare the inserting element with root, if less than root, then recurse for left, else recurse for right.

  3. After reaching the end, just insert that node at left(if less than current) else right.

Time Complexity: The worst-case time complexity of search and insert operations is O(h) where h is the height of the Binary Search Tree. In the worst case, we may have to travel from root to the deepest leaf node. The height of a skewed tree may become n and the time complexity of search and insert operation may become O(n).

Insertion using loop:

****

Output

10 15 20 30 40 50 60

Some Interesting Facts:

Related Links:

****

****

BST

Pros of a BST

  • When balanced, a BST provides lightning-fast O(log(n)) insertions, deletions, and lookups.

  • Binary search trees are pretty simple. An ordinary BST, unlike a balanced tree like a red-black tree, requires very little code to get running.

Cons of a BST

  • Slow for a brute-force search. If you need to iterate over each node, you might have more success with an array.

  • When the tree becomes unbalanced, all fast O(log(n)) operations quickly degrade to O(n).

  • Since pointers to whole objects are typically involved, a BST can require quite a bit more memory than an array, although this depends on the implementation.

Implementing a BST in Python

Step 1 – BSTNode Class

Our implementation won’t use a Tree class, but instead just a Node class. Binary trees are really just a pointer to a root node that in turn connects to each child node, so we’ll run with that idea.

First, we create a constructor:

class BSTNode:
    def __init__(self, val=None):
        self.left = None
        self.right = None
        self.val = valCode language: Python (python)

We’ll allow a value (key) to be provided, but if one isn’t provided we’ll just set it to None. We’ll also initialize both children of the new node to None.

Step 2 – Insert

We need a way to insert new data. The insert method is as follows:

def insert(self, val):
    if not self.val:
        self.val = val
        return

    if self.val == val:
        return

    if val < self.val:
        if self.left:
            self.left.insert(val)
            return
        self.left = BSTNode(val)
        return

    if self.right:
        self.right.insert(val)
        return
    self.right = BSTNode(val)Code language: Python (python)

If the node doesn’t yet have a value, we can just set the given value and return. If we ever try to insert a value that also exists, we can also simply return as this can be considered a noop. If the given value is less than our node’s value and we already have a left child then we recursively call insert on our left child. If we don’t have a left child yet then we just make the given value our new left child. We can do the same (but inverted) for our right side.

Step 3 – Get Min and Get Max

def get_min(self):
    current = self
    while current.left is not None:
        current = current.left
    return current.val

def get_max(self):
    current = self
    while current.right is not None:
        current = current.right
    return current.valCode language: Python (python)

getMin and getMax are useful helper functions, and they’re easy to write! They are simple recursive functions that traverse the edges of the tree to find the smallest or largest values stored therein.

Step 4 – Delete

def delete(self, val):
    if self == None:
        return self
    if val < self.val:
        self.left = self.left.delete(val)
        return self
    if val > self.val:
        self.right = self.right.delete(val)
        return self
    if self.right == None:
        return self.left
    if self.left == None:
        return self.right
    min_larger_node = self.right
    while min_larger_node.left:
        min_larger_node = min_larger_node.left
    self.val = min_larger_node.val
    self.right = self.right.delete(min_larger_node.val)
    return selfdef delete(self, val):
    if self == None:
        return self
    if val < self.val:
        if self.left:
            self.left = self.left.delete(val)
        return self
    if val > self.val:
        if self.right:
            self.right = self.right.delete(val)
        return self
    if self.right == None:
        return self.left
    if self.left == None:
        return self.right
    min_larger_node = self.right
    while min_larger_node.left:
        min_larger_node = min_larger_node.left
    self.val = min_larger_node.val
    self.right = self.right.delete(min_larger_node.val)
    return selfdef delete(self, val):
    if self == None:
        return self
    if val < self.val:
        self.left = self.left.delete(val)
        return self
    if val > self.val:
        self.right = self.right.delete(val)
        return self
    if self.right == None:
        return self.left
    if self.left == None:
        return self.right
    min_larger_node = self.right
    while min_larger_node.left:
        min_larger_node = min_larger_node.left
    self.val = min_larger_node.val
    self.right = self.right.delete(min_larger_node.val)
    return selfCode language: Python (python)

The delete operation is one of the more complex ones. It is a recursive function as well, but it also returns the new state of the given node after performing the delete operation. This allows a parent whose child has been deleted to properly set it’s left or right data member to None.

Step 5 – Exists

The exists function is another simple recursive function that returns True or False depending on whether a given value already exists in the tree.

def exists(self, val):
    if val == self.val:
        return True

    if val < self.val:
        if self.left == None:
            return False
        return self.left.exists(val)

    if self.right == None:
        return False
    return self.right.exists(val)Code language: Python (python)

Step 6 – Inorder

It’s useful to be able to print out the tree in a readable format. The inorder method print’s the values in the tree in the order of their keys.

def inorder(self, vals):
    if self.left is not None:
        self.left.inorder(vals)
    if self.val is not None:
        vals.append(self.val)
    if self.right is not None:
        self.right.inorder(vals)
    return valsCode language: Python (python)

Step 7 – Preorder

def preorder(self, vals):
    if self.val is not None:
        vals.append(self.val)
    if self.left is not None:
        self.left.preorder(vals)
    if self.right is not None:
        self.right.preorder(vals)
    return valsCode language: Python (python)

Step 8 – Postorder

def postorder(self, vals):
    if self.left is not None:
        self.left.postorder(vals)
    if self.right is not None:
        self.right.postorder(vals)
    if self.val is not None:
        vals.append(self.val)
    return valsCode language: Python (python)

Using the BST

def main():
    nums = [12, 6, 18, 19, 21, 11, 3, 5, 4, 24, 18]
    bst = BSTNode()
    for num in nums:
        bst.insert(num)
    print("preorder:")
    print(bst.preorder([]))
    print("#")

    print("postorder:")
    print(bst.postorder([]))
    print("#")

    print("inorder:")
    print(bst.inorder([]))
    print("#")

    nums = [2, 6, 20]
    print("deleting " + str(nums))
    for num in nums:
        bst.delete(num)
    print("#")

    print("4 exists:")
    print(bst.exists(4))
    print("2 exists:")
    print(bst.exists(2))
    print("12 exists:")
    print(bst.exists(12))
    print("18 exists:")
    print(bst.exists(18))Code language: Python (python)

Full Binary Search Tree in Python

class BSTNode:
    def __init__(self, val=None):
        self.left = None
        self.right = None
        self.val = val

    def insert(self, val):
        if not self.val:
            self.val = val
            return

        if self.val == val:
            return

        if val < self.val:
            if self.left:
                self.left.insert(val)
                return
            self.left = BSTNode(val)
            return

        if self.right:
            self.right.insert(val)
            return
        self.right = BSTNode(val)

    def get_min(self):
        current = self
        while current.left is not None:
            current = current.left
        return current.val

    def get_max(self):
        current = self
        while current.right is not None:
            current = current.right
        return current.val

    def delete(self, val):
        if self == None:
            return self
        if val < self.val:
            if self.left:
                self.left = self.left.delete(val)
            return self
        if val > self.val:
            if self.right:
                self.right = self.right.delete(val)
            return self
        if self.right == None:
            return self.left
        if self.left == None:
            return self.right
        min_larger_node = self.right
        while min_larger_node.left:
            min_larger_node = min_larger_node.left
        self.val = min_larger_node.val
        self.right = self.right.delete(min_larger_node.val)
        return self

    def exists(self, val):
        if val == self.val:
            return True

        if val < self.val:
            if self.left == None:
                return False
            return self.left.exists(val)

        if self.right == None:
            return False
        return self.right.exists(val)

    def preorder(self, vals):
        if self.val is not None:
            vals.append(self.val)
        if self.left is not None:
            self.left.preorder(vals)
        if self.right is not None:
            self.right.preorder(vals)
        return vals

    def inorder(self, vals):
        if self.left is not None:
            self.left.inorder(vals)
        if self.val is not None:
            vals.append(self.val)
        if self.right is not None:
            self.right.inorder(vals)
        return vals

    def postorder(self, vals):
        if self.left is not None:
            self.left.postorder(vals)
        if self.right is not None:
            self.right.postorder(vals)
        if self.val is not None:
            vals.append(self.val)
        return valsCode language: Python (python)

Where would you use a binary search tree in real life?

There are many applications of binary search trees in real life, and one of the most common use-cases is in storing indexes and keys in a database. For example, in MySQL or PostgresQL when you create a primary key column, what you’re really doing is creating a binary tree where the keys are the values of the column, and those nodes point to database rows. This lets the application easily search database rows by providing a key. For example, getting a user record by the email primary key.

There are many applications of binary search trees in real life, and one of the most common use cases is storing indexes and keys in a database. For example, when you create a primary key column in MySQL or PostgresQL, you create a binary tree where the keys are the values of the column and the nodes point to database rows. This allows the application to easily search for database rows by specifying a key, for example, to find a user record using the email primary key.

Other common uses include:

  • Pathfinding algorithms in videogames (A*) use BSTs

  • File compression using a Huffman encoding scheme uses a binary search tree

  • Rendering calculations – Doom (1993) was famously the first game to use a BST

  • Compilers for low-level coding languages parse syntax using a BST

  • Almost every database in existence uses BSTs for key lookups

Example Binary Tree

Visual Aid

Example Code

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

let a = new TreeNode("a");
let b = new TreeNode("b");
let c = new TreeNode("c");
let d = new TreeNode("d");
let e = new TreeNode("e");
let f = new TreeNode("f");

a.left = b;
a.right = c;
b.left = d;
b.right = e;
c.right = f;

Terms

  • tree - graph with no cycles

  • binary tree - tree where nodes have at most 2 nodes

  • root - the ultimate parent, the single node of a tree that can access every other node through edges; by definition the root will not have a parent

  • internal node - a node that has children

  • leaf - a node that does not have any children

  • path - a series of nodes that can be traveled through edges - for example A, B, E is a path through the above tree

Search Patterns

  • Breadth First Search - Check all nodes at a level before moving down a level

    • Think of this of searching horizontally in rows

  • Depth First Search - Check the depth as far as it goes for one child, before

    moving on to the next child.

    • Think of this as searching vertically in diagonals

    • Pre-Order Traversal - Access the data of the current node, recursively visit the left sub tree, recursively visit the right sub tree

      • All the way to the left, top down, going right after other options have already been logged.

    • In-Order Traversal - Recursively visit the left sub tree, access the data of the current node, recursively visit the right sub tree

      • In the order they were the "current root", the actual return order of the recursive calls.

    • Post-Order Traversal - Recursively visit the left sub tree, recursively visit the right sub tree, access the data of the current node.

      • Starting with the bottom most nodes working up through the tree

Constraints

  • Binary trees have at most two children per node

  • Given any node of the tree, the values on the left must be strictly less than that node

  • Given any node of the tree, the values on the right must be strictly greater than or equal to that node

  • Given these constraints a binary tree is necessarily a sorted data structure

  • The worst binary trees devolve into a linked list, the best are height balanced (think branching).

PseudoCode For Insertion

  • Create a new node

  • Start at the root

    • Check if there is a root

      • If not the root becomes the new node

    • If there is a root check if the value of the new node is equal to, greater then, or less then the value of the root

      • If it is greater or equal to

        • Check to see if there is a node to the right

          • If there is, move to the new node and continue with the node to the right as the subtree root

          • If there is not, add the new node as the right property of the current node

      • If it is smaller

        • Check to see if there is a node to the left

          • If there is, move to the new node and continue with the node to the left as the subtree root

          • If there is not, add the new node as the left property of the current node

PseudoCode For Search Of A single Item

  • Start at the root

    • Check if there is a root

      • If not, there is nothing in the tree, so the search is over

    • If there is a root, check if the value of the root is equal to, greater then, or less then the value were looking for;

      • If it is equal to the value

        • We found what we are searching for

      • If it is less than the value

        • Check to see if there is a node to the left

          • If there isn't

            • the value isn't in our tree

          • If there is

            • repeat these steps with the node to the left as the new subtree root

      • If it is greater than the value

        • Check to see if there is a node to the right

          • If there isn't

            • the value isn't in our tree

          • If there is

            • repeat these steps with the node to the right as the new subtree root

PseudoCode For Breadth First Search Traversal

  • Create a queue class or use an array

  • Create a variable to store the values of the nodes visited

  • Place the root in the queue

  • Loop as many times as there are items in the queue

    • Dequeue a node

    • If there is a left value to the node dequeued, add it to the queue

    • If there is a right value to the node dequeued, add it to the queue

    • Push the nodes value into the variable that stores nodes visited

PseudoCode For Depth First Search Traversal

Pre-Order

Iterative

  • Create a stack class or use an array

  • Push the root into the stack

  • Create a variable to store the values of the nodes visited

  • Do this as long as there is something on the stack

    • Pop a node from the stack

    • Push that nodes value into the variable that stores nodes visited.

    • If there is a node to the right push it into the stack

    • If there is a node to the left push it into the stack

  • Return the variable storing the values

Recursive

  • Create a variable to store the current root

  • Push the value of current root to the variable storing the values

  • If the current root has a left propety call the function on that the left property

  • If the current root has a right propety call the function on that the right property

  • Spread the current root, the left values, and the right values

In-Order

Iterative

  • Create a stack class or use an array

  • Create a variable to store the current root

  • Create a variable to store the values of the nodes visited

  • Create a loop

    • While the current root exists

      • push the current root to the call stack

      • current root is equal to the left of current root

    • if the stack is empty break out of the loop

    • set a variable to equal the popped value of the stack

    • push that variable into the variable that stores values

    • set the current root to the right of the current loop

  • Return the variable storing the values

Recursive

  • Create a variable to store the current root

  • Push the value of current root to the variable storing the values

  • If the current root has a left propety call the function on that the left property

  • If the current root has a right propety call the function on that the right property

  • Spread the the left values, current root ,and the right values

Post-Order

Iterative

  • Haven't figured this one out yet.

Recursive

  • Create a variable to store the current root

  • Push the value of current root to the variable storing the values

  • If the current root has a left propety call the function on that the left property

  • If the current root has a right propety call the function on that the right property

  • Spread the the left values, the right values, and current root

Example Binary Search Tree

class TreeNode {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

class BST {
  constructor() {
    this.root = null;
  }

  //Insert a new node

  recursiveInsert(val, currentNode = this.root) {
    if (!this.root) {
      this.root = new TreeNode(val);
      return this;
    }
    if (val < currentNode.val) {
      if (!currentNode.left) {
        currentNode.left = new TreeNode(val);
      } else {
        this.insert(val, currentNode.left);
      }
    } else {
      if (!currentNode.right) {
        currentNode.right = new TreeNode(val);
      } else {
        this.insert(val, currentNode.right);
      }
    }
  }

  iterativeInsert(val, currentNode = this.root) {
    if (!this.root) {
      this.root = new TreeNode(val);
      return this;
    }
    if (val < currentNode.val) {
      if (!currentNode.left) {
        currentNode.left = new TreeNode();
      } else {
        while (true) {
          if (val < currentNode.val) {
            if (!currenNodet.left) {
              currentNode.left = new TreeNode();
              return this;
            } else {
              currentNode = currentNode.left;
            }
          } else {
            if (!currentNode.right) {
              currentNode.right = new TreeNode();
              return this;
            } else {
              currentNode = currentNode.right;
            }
          }
        }
      }
    }
  }

  //Search the tree

  searchRecur(val, currentNode = this.root) {
    if (!currentNode) return false;
    if (val < currentNode.val) {
      return this.searchRecur(val, currentNode.left);
    } else if (val > currentNode.val) {
      return this.searchRecur(val, currentNode.right);
    } else {
      return true;
    }
  }

  searchIter(val) {
    let currentNode = this.root;
    while (currentNode) {
      if (val < currentNode.val) {
        currentNode = currentNode.left;
      } else if (val > currentNode.val) {
        currentNode = currentNode.right;
      } else {
        return true;
      }
    }
    return false;
  }

  // Maybe works, who knows, pulled it off the internet....

  deleteNodeHelper(root, key) {
    if (root === null) {
      return false;
    }
    if (key < root.val) {
      root.left = deleteNodeHelper(root.left, key);
      return root;
    } else if (key > root.val) {
      root.right = deleteNodeHelper(root.right, key);
      return root;
    } else {
      if (root.left === null && root.right === null) {
        root = null;
        return root;
      }
      if (root.left === null) return root.right;
      if (root.right === null) return root.left;

      let currNode = root.right;
      while (currNode.left !== null) {
        currNode = currNode.left;
      }
      root.val = currNode.val;
      root.right = deleteNodeHelper(root.right, currNode.val);
      return root;
    }
  }

  //Recursive Depth First Search

  preOrderTraversal(root) {
    if (!root) return [];
    let left = this.preOrderTraversal(root.left);
    let right = this.preOrderTraversal(root.right);
    return [root.val, ...left, ...right];
  }

  preOrderTraversalV2(root) {
    if (!root) return [];
    let newArray = new Array();
    newArray.push(root.val);
    newArray.push(...this.preOrderTraversalV2(root.left));
    newArray.push(...this.preOrderTraversalV2(root.right));
    return newArray;
  }

  inOrderTraversal(root) {
    if (!root) return [];
    let left = this.inOrderTraversal(root.left);
    let right = this.inOrderTraversal(root.right);
    return [...left, root.val, ...right];
  }

  inOrderTraversalV2(root) {
    if (!root) return [];
    let newArray = new Array();
    newArray.push(...this.inOrderTraversalV2(root.left));
    newArray.push(root.val);
    newArray.push(...this.inOrderTraversalV2(root.right));
    return newArray;
  }

  postOrderTraversal(root) {
    if (!root) return [];
    let left = this.postOrderTraversal(root.left);
    let right = this.postOrderTraversal(root.right);
    return [...left, ...right, root.val];
  }

  postOrderTraversalV2(root) {
    if (!root) return [];
    let newArray = new Array();
    newArray.push(...this.postOrderTraversalV2(root.left));
    newArray.push(...this.postOrderTraversalV2(root.right));
    newArray.push(root.val);
    return newArray;
  }

  // Iterative Depth First Search

  iterativePreOrder(root) {
    let stack = [root];
    let results = [];
    while (stack.length) {
      let current = stack.pop();
      results.push(current);
      if (current.left) stack.push(current.left);
      if (current.right) stack.push(current.right);
    }
    return results;
  }

  iterativeInOrder(root) {
    let stack = [];
    let current = root;
    let results = [];
    while (true) {
      while (current) {
        stack.push(current);
        current = current.left;
      }

      if (!stack.length) break;
      let removed = stack.pop();
      results.push(removed);
      current = current.right;
    }
    return results;
  }

  //To-Do iterativePostOrder

  //Breadth First Search

  breadthFirstSearch(root) {
    let queue = [root];
    let result = [];
    while (queue.length) {
      let current = queue.shift();
      if (current.left) queue.push(current.left);
      if (current.right) queue.push(current.left);
      current.push(result);
    }
    return result;
  }

  // Converting a Sorted Array to a Binary Search Tree

  sortedArrayToBST(nums) {
    if (nums.length === 0) return null;

    let mid = Math.floor(nums.length / 2);
    let root = new TreeNode(nums[mid]);

    let left = nums.slice(0, mid);
    root.left = sortedArrayToBST(left);

    let right = nums.slice(mid + 1);
    root.right = sortedArrayToBST(right);

    return root;
  }
}
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