Binary Search Trees

13 minute read

The Binary Search Tree (BST) is a Binary Tree with the following properties.

  1. Keys that are less than the parent are found in the left subtree
  2. Keys that are greater than the parent are found in the right subtree
  3. Both the left and right subtrees must also be binary search trees.

Binary Search Tree

Binary Search Tree

Binary Search Tree Operations

Operation Average Worst Case
Insert O(log n) O(n)
Lookup O(log n) O(n)
Delete O(log n) O(n)
Minimum O(log n) O(n)
Maximum O(log n) O(n)
Predecessor O(log n) O(n)
Successor O(log n) O(n)

When we are talking about the average case, it is the time it takes for the operation on a balanced tree, and when we are talking about the worst case, it is the time it takes for the given operation on a non-balanced tree.

Balanced and Non-Balanced Trees


We should start the implementation by defining BinarySearchTree() and Node() classes in our code.

An empty BinarySearchTree() class will look like this;

class BinarySearchTree:

    def __init__(self):
        self.root = None
        self.size = 0

    def length(self):
        return self.size

    def __len__(self):
        return self.size

and along with the Node class, we will implement lots of helper functions like; has_left_child(), has_right_child(), is_left_child(), is_right_child(), is_root(), is_leaf(), has_any_children(), has_both_children(), splice_out(), find_successor(), find_min(), replace_node_data() in order to make our future implementations more easier.

class Node:
    def __init__(self, key, val, left=None, right=None, parent=None):
        self.key = key
        self.payload = val
        self.leftChild = left
        self.rightChild = right
        self.parent = parent

    def has_left_child(self):
        return self.leftChild

    def has_right_child(self):
        return self.rightChild

    def is_left_child(self):
        return self.parent and self.parent.leftChild == self

    def is_right_child(self):
        return self.parent and self.parent.rightChild == self

    def is_root(self):
        return not self.parent

    def is_leaf(self):
        return not (self.rightChild or self.leftChild)

    def has_any_children(self):
        return self.rightChild or self.leftChild

    def has_both_children(self):
        return self.rightChild and self.leftChild

Insert Operation

Since we now have a Node() and BinarySearchTree() classes, we are ready to insert elements to this BinarySearchTree() class.

We are going to implement a put(self, key, val) method. This method will check to see if the tree already has a root. If there is not a root then put() will create a new Node() and install it as the root of the tree. If a root node is already in place then put() calls the private, recursive, helper function _put() to search the tree according to the Binary Search Tree properties that we explained in the first paragraph of this article.

def put(self, key, val):
    if self.root:
        self._put(key, val, self.root)
        self.root = Node(key, val)
    self.size = self.size + 1

def _put(self, key, val, current_node):
    if key < current_node.key:
        if current_node.has_left_child():
            self._put(key, val, current_node.leftChild)
            current_node.leftChild = Node(key, val, parent=current_node)
        if current_node.has_right_child():
            self._put(key, val, current_node.rightChild)
            current_node.rightChild = Node(key, val, parent=current_node)

def __setitem__(self, k, v):
    self.put(k, v)

Lookup (Search) Operation

Once the tree is constructed, the next task is to implement the retrieval of a value for a given key. The get() method is even easier than the put() method because it simply searches the tree recursively until it gets to a non-matching leaf node or finds a matching key. When a matching key is found, the value stored in the payload of the node is returned.

def get(self, key):
    if self.root:
        result = self._get(key, self.root)
        if result:
            return result.payload
            return None
        return None

def _get(self, key, current_node):
    if not current_node:
        return None
    elif current_node.key == key:
        return current_node
    elif key < current_node.key:
        return self._get(key, current_node.leftChild)
        return self._get(key, current_node.rightChild)

def __getitem__(self, key):
    return self.get(key)

def __contains__(self, key):
    if self._get(key, self.root):
        return True
        return False

Delete Operation

delete() operation is the most challenging operation in the Binary Search Tree.

The first task is to find the node to delete by searching the tree. If the tree has more than one node we search using the _get() method to find the Node() that needs to be removed. If the tree only has a single node, that means we are removing the root of the tree, but we still must check to make sure the key of the root matches the key that is to be deleted.

In either case if the key is not found the delete() method raises an error.

def delete(self, key):
    if self.size > 1:
        node_to_remove = self._get(key, self.root)
        if node_to_remove:
            self.size = self.size - 1
            raise KeyError('Error, key not in tree')
    elif self.size == 1 and self.root.key == key:
        self.root = None
        self.size = self.size - 1
        raise KeyError('Error, key not in tree')

def __delitem__(self, key):

Once we have found the node containing the key we want to delete, there are three cases that we must consider:

  1. The node to be deleted has no children
  2. The node to be deleted has only one child
  3. The node to be deleted has two children

Handling the first case is pretty easy:

Deleting a Node Without Children

Deleting Node 65, a node without children
def remove(self, current_node):
    if current_node.is_leaf():  # leaf
        if current_node == current_node.parent.leftChild:
            current_node.parent.leftChild = None
            current_node.parent.rightChild = None

If a node has only a single child, then we can simply promote the child to take the place of its parent.

Deleting a Node With Single Children

Deleting Node 89, a node with single children

The decision proceeds as follows:

  1. If the current node is a left child then we only need to update the parent reference of the left child to point to the parent of the current node, and then update the left child reference of the parent to point to the current node’s left child.
  2. If the current node is a right child then we only need to update the parent reference of the left child to point to the parent of the current node, and then update the right child reference of the parent to point to the current node’s left child.
  3. If the current node has no parent, it must be the root. In this case we will just replace the key, payload, leftChild, and rightChild data by calling the replace_node_data() method on the root.
else:  # this node has one child
    if current_node.has_left_child():
        if current_node.is_left_child():
            current_node.leftChild.parent = current_node.parent
            current_node.parent.leftChild = current_node.leftChild
        elif current_node.is_right_child():
            current_node.leftChild.parent = current_node.parent
            current_node.parent.rightChild = current_node.leftChild
        if current_node.is_left_child():
            current_node.rightChild.parent = current_node.parent
            current_node.parent.leftChild = current_node.rightChild
        elif current_node.is_right_child():
            current_node.rightChild.parent = current_node.parent
            current_node.parent.rightChild = current_node.rightChild

If a node has two children, then it is unlikely that we can simply promote one of them to take the node’s place. We can, however, search the tree for a node that can be used to replace the one scheduled for deletion.

What we need is a node that will preserve the binary search tree relationships for both of the existing left and right subtrees. The node that will do this is the node that has the next-largest key in the tree. We call this node the successor, and we will look at a way to find the successor shortly.

The successor is guaranteed to have no more than one child, so we know how to remove it using the two cases for deletion that we have already implemented.

Deleting a Node With Two Children

Deleting Node 18, a node with two children
elif current_node.has_both_children():
    successor = current_node.find_successor()
    current_node.key = successor.key
    current_node.payload = successor.payload

There are three cases to consider when looking for the successor:

  1. If the node has a right child, then the successor is the smallest key in the right subtree
  2. If the node has no right child and is the left child of its parent, then the parent is the successor
  3. If the node is the right child of its parent, and itself has no right child, then the successor to this node is the successor of its parent, excluding this node.
def splice_out(self):
    if self.is_leaf():
        if self.is_left_child():
            self.parent.leftChild = None
            self.parent.rightChild = None
    elif self.has_any_children():
        if self.has_left_child():
            if self.is_left_child():
                self.parent.leftChild = self.leftChild
                self.parent.rightChild = self.leftChild
            self.leftChild.parent = self.parent
            if self.is_left_child():
                self.parent.leftChild = self.rightChild
                self.parent.rightChild = self.rightChild
            self.rightChild.parent = self.parent

def find_successor(self):
    successor = None
    if self.has_right_child():
        successor = self.rightChild.find_min()
        if self.parent:
            if self.is_left_child():
                successor = self.parent
                self.parent.rightChild = None
                successor = self.parent.find_successor()
                self.parent.rightChild = self
    return successor

def find_min(self):
    current = self
    while current.has_left_child():
        current = current.leftChild
    return current

def replace_node_data(self, key, value, lc, rc):
    self.key = key
    self.payload = value
    self.leftChild = lc
    self.rightChild = rc
    if self.has_left_child():
        self.leftChild.parent = self
    if self.has_right_child():
        self.rightChild.parent = self

Full Implementation Code

Full implementation code can be found at my GitHub page.