Introduction to Binary Search Tree – Data Structure and Algorithm Tutorials

Last Updated : 16 Apr, 2024

Binary Search Tree (BST) is a special type of binary tree in which the left child of a node has a value less than the node’s value and the right child has a value greater than the node’s value. This property is called the BST property and it makes it possible to efficiently search, insert, and delete elements in the tree.

Properties of Binary Search Tree:

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.
This means everything to the left of the root is less than the value of the root and everything to the right of the root is greater than the value of the root. Due to this performing, a binary search is very easy.
The left and right subtree each must also be a binary search tree.
There must be no duplicate nodes(BST may have duplicate values with different handling approaches)

Binary Search Tree

Handling duplicate values in the Binary Search Tree:

We must follow a consistent process throughout i.e. either store duplicate value at the left or store the duplicate value at the right of the root, but be consistent with your approach.

Operations performed on a BST:

1. Insert a node into a BST:

A new key is always inserted at the leaf. Start searching a key from the root till a leaf node. Once a leaf node is found, the new node is added as a child of the leaf node.

C++

// Given Node Structure
struct node
{
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp = (struct node*)malloc(
                  sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key)
    {
        node->left = insert(node->left, key);
    }
    else if (key > node->key)
    {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

// Given Node Structure
struct node {
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key) {
        node->left = insert(node->left, key);
    }
    else if (key > node->key) {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

Java

class GFG {

    // Given Node Structure
    static class node {
        int key;
        node left, right;
    };

    // Function to create a new BST node
    static node newNode(int item)
    {
        node temp = new node();
        temp.key = item;
        temp.left = temp.right = null;
        return temp;
    }

    // Function to insert a new node with
    // given key in BST
    static node insert(node node, int key)
    {
        // If the tree is empty, return a new node
        if (node == null)
            return newNode(key);

        // Otherwise, recur down the tree
        if (key < node.key) {
            node.left = insert(node.left, key);
        }
        else if (key > node.key) {
            node.right = insert(node.right, key);
        }

        // Return the node
        return node;
    }
}

Python3

# Given Node Structure
class Node:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None

# Function to insert a new node with
# given key in BST
def insert(node, key):
    # If the tree is empty, return a new node
    if node is None:
        return Node(key)

    # Otherwise, recur down the tree
    if key < node.key:
        node.left = insert(node.left, key)
    elif key > node.key:
        node.right = insert(node.right, key)

    # Return the node pointer
    return node

Javascript

// Given Node Structure
class Node
{
    constructor(key){
        this.key = key;
        this.left = null;
        this.right = null;
    }
}

// Function to insert a new node with
// given key in BST
function insert(node, key)
{
    
    // If the tree is empty, return a new node
    if (node == null)
        return new Node(key);

    // Otherwise, recur down the tree
    if (key < node.key)
    {
        node.left = insert(node.left, key);
    }
    else if (key > node.key)
    {
        node.right = insert(node.right, key);
    }

    // Return the node pointer
    return node;
}

Time Complexity: O(N), where N is the number of nodes of the BST
Auxiliary Space: O(1)

2. Delete a Node of BST: It is used to delete a node with specific key from the BST and return the new BST.

Different scenarios for deleting the node:

Node to be deleted is the leaf node : Its simple you can just null it out.
Node to be deleted has one child : You can just replace the node with the child node.
Node to be deleted has two child : Here we have to delete the node is such a way, that the resulting tree follows the properties of a BST. The trick is to find the inorder successor of the node. Copy contents of the inorder successor to the node, and delete the inorder successor.

Take Care of following things while deleting a node of a BST:

Need to figure out what will be the replacement of the node to be deleted.
Want minimal disruption to the existing tree structure
Can take the replacement node from the deleted nodes left or right subtree.
If taking if from the left subtree, we have to take the largest value in the left subtree.
If taking if from the right subtree, we have to take the smallest value in the right subtree.

Code:

Below is the implementation of the deletion in BST:

C++

// C++ program to delete
// a node of BST
// Given Node node
struct node {
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key) {
        node->left = insert(node->left, key);
    }
    else if (key > node->key) {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

// Function that returns the node with minimum
// key value found in that tree
struct node* minValueNode(struct node* node)
{
    struct node* current = node;

    // Loop down to find the leftmost leaf
    while (current && current->left != NULL)
        current = current->left;

    return current;
}

// Function that deletes the key and
// returns the new root
struct node* deleteNode(struct node* root,
                        int key)
{
    // base Case
    if (root == NULL)
        return root;

    // If the key to be deleted is
    // smaller than the root's key,
    // then it lies in left subtree
    if (key < root->key) {
        root->left
            = deleteNode(root->left, key);
    }

    // If the key to be deleted is
    // greater than the root's key,
    // then it lies in right subtree
    else if (key > root->key) {

        root->right
            = deleteNode(root->right, key);
    }

    // If key is same as root's key,
    // then this is the node
    // to be deleted
    else {

        // Node with only one child
        // or no child
        if (root->left == NULL) {
            struct node* temp = root->right;
            free(root);
            return temp;
        }
        else if (root->right == NULL) {
            struct node* temp = root->left;
            free(root);
            return temp;
        }

        // Node with two children:
        // Get the inorder successor(smallest
        // in the right subtree)
        struct node* temp = minValueNode(root->right);

        // Copy the inorder successor's
        // content to this node
        root->key = temp->key;

        // Delete the inorder successor
        root->right
            = deleteNode(root->right, temp->key);
    }
    return root;
}

// C program to delete
// a node of BST
// Given Node node
struct node {
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key) {
        node->left = insert(node->left, key);
    }
    else if (key > node->key) {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

// Function that returns the node with minimum
// key value found in that tree
struct node* minValueNode(struct node* node)
{
    struct node* current = node;

    // Loop down to find the leftmost leaf
    while (current && current->left != NULL)
        current = current->left;

    return current;
}

// Function that deletes the key and
// returns the new root
struct node* deleteNode(struct node* root,
                        int key)
{
    // base Case
    if (root == NULL)
        return root;

    // If the key to be deleted is
    // smaller than the root's key,
    // then it lies in left subtree
    if (key < root->key) {
        root->left
            = deleteNode(root->left, key);
    }

    // If the key to be deleted is
    // greater than the root's key,
    // then it lies in right subtree
    else if (key > root->key) {

        root->right
            = deleteNode(root->right, key);
    }

    // If key is same as root's key,
    // then this is the node
    // to be deleted
    else {

        // Node with only one child
        // or no child
        if (root->left == NULL) {
            struct node* temp = root->right;
            free(root);
            return temp;
        }
        else if (root->right == NULL) {
            struct node* temp = root->left;
            free(root);
            return temp;
        }

        // Node with two children:
        // Get the inorder successor(smallest
        // in the right subtree)
        struct node* temp = minValueNode(root->right);

        // Copy the inorder successor's
        // content to this node
        root->key = temp->key;

        // Delete the inorder successor
        root->right
            = deleteNode(root->right, temp->key);
    }
    return root;
}

Java

// Java program for Delete a Node of BST
class GFG {

    // Given Node node
    static class node {
        int key;
        node left, right;
    };

    // Function to create a new BST node
    static node newNode(int item)
    {
        node temp = new node();
        temp.key = item;
        temp.left = temp.right = null;
        return temp;
    }

    // Function to insert a new node with
    // given key in BST
    static node insert(node node, int key)
    {
        // If the tree is empty, return a new node
        if (node == null)
            return newNode(key);

        // Otherwise, recur down the tree
        if (key < node.key) {
            node.left = insert(node.left, key);
        }
        else if (key > node.key) {
            node.right = insert(node.right, key);
        }

        // Return the node
        return node;
    }
    // Function that returns the node with minimum
    // key value found in that tree
    static node minValueNode(node node)
    {
        node current = node;

        // Loop down to find the leftmost leaf
        while (current != null && current.left != null)
            current = current.left;

        return current;
    }

    // Function that deletes the key and
    // returns the new root
    static node deleteNode(node root, int key)
    {
        // base Case
        if (root == null)
            return root;

        // If the key to be deleted is
        // smaller than the root's key,
        // then it lies in left subtree
        if (key < root.key) {
            root.left = deleteNode(root.left, key);
        }

        // If the key to be deleted is
        // greater than the root's key,
        // then it lies in right subtree
        else if (key > root.key) {

            root.right = deleteNode(root.right, key);
        }

        // If key is same as root's key,
        // then this is the node
        // to be deleted
        else {

            // Node with only one child
            // or no child
            if (root.left == null) {
                node temp = root.right;
                return temp;
            }
            else if (root.right == null) {
                node temp = root.left;
                return temp;
            }

            // Node with two children:
            // Get the inorder successor(smallest
            // in the right subtree)
            node temp = minValueNode(root.right);

            // Copy the inorder successor's
            // content to this node
            root.key = temp.key;

            // Delete the inorder successor
            root.right = deleteNode(root.right, temp.key);
        }
        return root;
    }

Python

# Python program to delete a node of BST

# Given Node node
class Node:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None

# Function to insert a new node with
# given key in BST
def insert(root, key):
    # If the tree is empty, return a new node
    if root is None:
        return Node(key)

    # Otherwise, recur down the tree
    if key < root.key:
        root.left = insert(root.left, key)
    elif key > root.key:
        root.right = insert(root.right, key)

    # Return the node pointer
    return root

# Function to do inorder traversal of BST
def inorder(root):
    if root:
        inorder(root.left)
        print(root.key, end=" ")
        inorder(root.right)

# Function that returns the node with minimum
# key value found in that tree
def minValueNode(node):
    current = node

    # Loop down to find the leftmost leaf
    while current and current.left is not None:
        current = current.left

    return current

# Function that deletes the key and
# returns the new root
def deleteNode(root, key):
    # base Case
    if root is None:
        return root

    # If the key to be deleted is
    # smaller than the root's key,
    # then it lies in left subtree
    if key < root.key:
        root.left = deleteNode(root.left, key)

    # If the key to be deleted is
    # greater than the root's key,
    # then it lies in right subtree
    elif key > root.key:

        root.right = deleteNode(root.right, key)

    # If key is same as root's key,
    # then this is the node
    # to be deleted
    else:

        # Node with only one child
        # or no child
        if root.left is None:
            temp = root.right
            root = None
            return temp
        elif root.right is None:
            temp = root.left
            root = None
            return temp

        # Node with two children:
        # Get the inorder successor(smallest
        # in the right subtree)
        temp = minValueNode(root.right)

        # Copy the inorder successor's
        # content to this node
        root.key = temp.key

        # Delete the inorder successor
        root.right = deleteNode(root.right, temp.key)

    return root

3. Traversal (Inorder traversal of BST) : In case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. We visit the left child first, then the root, and then the right child.

Below is the implementation of how to do inorder traversal of a Binary Search Tree:

C++

// C++ program to implement
// inorder traversal of BST
#include <bits/stdc++.h>
using namespace std;

// Given Node node
struct node
{
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp = (struct node*)malloc(
            sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key) 
    {
        node->left = insert(node->left, key);
    }
    else if (key > node->key)
    {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

// Function to do inorder traversal of BST
void inorder(struct node* root)
{
    if (root != NULL) 
    {
        inorder(root->left);
        cout << " " << root->key;
        inorder(root->right);
    }
}

// Driver Code
int main()
{
    
    /* Let us create following BST 
              50 
           /     \ 
          30      70 
         /  \    /  \ 
       20   40  60   80 
   */
    struct node* root = NULL;

    // Creating the BST
    root = insert(root, 50);
    insert(root, 30);
    insert(root, 20);
    insert(root, 40);
    insert(root, 70);
    insert(root, 60);
    insert(root, 80);

    // Function Call
    inorder(root);

    return 0;
}

// This code is contributed by shivanisinghss2110

// C program to implement
// inorder traversal of BST
#include <stdio.h>
#include <stdlib.h>

// Given Node node
struct node {
    int key;
    struct node *left, *right;
};

// Function to create a new BST node
struct node* newNode(int item)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// Function to insert a new node with
// given key in BST
struct node* insert(struct node* node, int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);

    // Otherwise, recur down the tree
    if (key < node->key) {
        node->left = insert(node->left, key);
    }
    else if (key > node->key) {
        node->right = insert(node->right, key);
    }

    // Return the node pointer
    return node;
}

// Function to do inorder traversal of BST
void inorder(struct node* root)
{
    if (root != NULL) {
        inorder(root->left);
        printf("%d ", root->key);
        inorder(root->right);
    }
}

// Driver Code
int main()
{
    /* Let us create following BST 
              50 
           /     \ 
          30      70 
         /  \    /  \ 
       20   40  60   80 
   */
    struct node* root = NULL;

    // Creating the BST
    root = insert(root, 50);
    insert(root, 30);
    insert(root, 20);
    insert(root, 40);
    insert(root, 70);
    insert(root, 60);
    insert(root, 80);

    // Function Call
    inorder(root);

    return 0;
}

Java

import java.io.*;

// Java program for Inorder Traversal
class GFG {

    // Given Node node
    static class node {
        int key;
        node left, right;
    };

    // Function to create a new BST node
    static node newNode(int item)
    {
        node temp = new node();
        temp.key = item;
        temp.left = temp.right = null;
        return temp;
    }

    // Function to insert a new node with
    // given key in BST
    static node insert(node node, int key)
    {
        // If the tree is empty, return a new node
        if (node == null)
            return newNode(key);

        // Otherwise, recur down the tree
        if (key < node.key) {
            node.left = insert(node.left, key);
        }
        else if (key > node.key) {
            node.right = insert(node.right, key);
        }

        // Return the node
        return node;
    }

    // Function to do inorder traversal of BST
    static void inorder(node root)
    {
        if (root != null) {
            inorder(root.left);
            System.out.print(" " + root.key);
            inorder(root.right);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {

        /* Let us create following BST
                50
             /     \
            30      70
           /  \    /  \
         20   40  60   80
     */
        node root = null;

        // inserting value 50
        root = insert(root, 50);

        // inserting value 30
        insert(root, 30);

        // inserting value 20
        insert(root, 20);

        // inserting value 40
        insert(root, 40);

        // inserting value 70
        insert(root, 70);

        // inserting value 60
        insert(root, 60);

        // inserting value 80
        insert(root, 80);

        // print the BST
        inorder(root);
    }
}
// This code is contributed by abhijitjadhav1998

Python3

# Python program to implement
# inorder traversal of BST

# Given Node node
class Node:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None

# Function to create a new BST node
def newNode(item):
    temp = Node(item)
    temp.key = item
    temp.left = temp.right = None
    return temp

# Function to insert a new node with
# given key in BST
def insert(node, key):
    # If the tree is empty, return a new node
    if node is None:
        return newNode(key)

    # Otherwise, recur down the tree
    if key < node.key:
        node.left = insert(node.left, key)
    elif key > node.key:
        node.right = insert(node.right, key)

    # Return the node pointer
    return node

# Function to do inorder traversal of BST
def inorder(root):
    if root:
        inorder(root.left)
        print(root.key, end=" ")
        inorder(root.right)

# Driver Code
if __name__ == '__main__':
    
    # Let us create following BST 
    #          50 
    #       /     \ 
    #     30      70 
    #    /  \    /  \ 
    #  20   40  60   80 
    root = None

    # Creating the BST
    root = insert(root, 50)
    insert(root, 30)
    insert(root, 20)
    insert(root, 40)
    insert(root, 70)
    insert(root, 60)
    insert(root, 80)

    # Function Call
    inorder(root)
#This code is contributed by japmeet01

Output

 20 30 40 50 60 70 80

Time Complexity: O(N), where N is the number of nodes of the BST
Auxiliary Space: O(1)

Applications of BST:

Graph algorithms: BSTs can be used to implement graph algorithms, such as in minimum spanning tree algorithms.
Priority Queues: BSTs can be used to implement priority queues, where the element with the highest priority is at the root of the tree, and elements with lower priority are stored in the subtrees.
Self-balancing binary search tree: BSTs can be used as a self-balancing data structures such as AVL tree and Red-black tree.
Data storage and retrieval: BSTs can be used to store and retrieve data quickly, such as in databases, where searching for a specific record can be done in logarithmic time.

Advantages:

Fast search: Searching for a specific value in a BST has an average time complexity of O(log n), where n is the number of nodes in the tree. This is much faster than searching for an element in an array or linked list, which have a time complexity of O(n) in the worst case.
In-order traversal: BSTs can be traversed in-order, which visits the left subtree, the root, and the right subtree. This can be used to sort a dataset.
Space efficient: BSTs are space efficient as they do not store any redundant information, unlike arrays and linked lists.

Disadvantages:

Skewed trees: If a tree becomes skewed, the time complexity of search, insertion, and deletion operations will be O(n) instead of O(log n), which can make the tree inefficient.
Additional time required: Self-balancing trees require additional time to maintain balance during insertion and deletion operations.
Efficiency: BSTs are not efficient for datasets with many duplicates as they will waste space.

Suggest improvement

Binary Search Tree

Applications of BST

Share your thoughts in the comments

Easy problems on Binary Search Tree

Intermediate problems on Binary Search Tree

Hard problems on Binary Search Tree

Easy problems on Binary Search Tree

Intermediate problems on Binary Search Tree

Hard problems on Binary Search Tree

Introduction to Binary Search Tree – Data Structure and Algorithm Tutorials

Properties of Binary Search Tree:

Handling duplicate values in the Binary Search Tree:

Operations performed on a BST:

Applications of BST:

Advantages:

Disadvantages:

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