Insertion in Binary Search Tree
What is a Binary Search Tree?
A binary Search Tree is a special type of binary tree data structure that 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.
Binary Search Tree
The above properties of a Binary Search Tree provide an ordering among keys so that the operations like search, minimum and maximum can be done fast.
Insert a value in a Binary Search Tree:
A new key is always inserted at the leaf by maintaining the property of the binary search tree. We start searching for 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. The below steps are followed while we try to insert a node into a binary search tree:
- Check the value to be inserted (say X) with the value of the current node (say val) we are in:
- If X is less than val move to the left subtree.
- Otherwise, move to the right subtree.
- Once the leaf node is reached, insert X to its right or left based on the relation between X and the leaf node’s value.
Follow the below illustration for a better understanding:
Illustration:
Consider the below tree:
Binary Search Tree
Let us try to insert a node with value 40 in this tree:
1st step: 40 will be compared with root, i.e., 100.
- 40 is less than 100.
- So move to the left subtree of 100. The root of the left subtree is 20.
2nd step: 40 is now compared with 20.
- It is greater than 20.
- So move to the right subtree of 20 whose root is 30.
3rd step: 30 is a leaf node.
- So we have to insert 40 to the left or right of 30.
- As 40 is greater than 30, insert 40 to the right of 30.
The new tree will look like the following:
Insertion in Binary Search Tree
Example no 1 – Implementation using Recursion:
Below is the implementation of the insertion operation using recursion.
C
// C program to demonstrate insert
// operation in binary
// search tree.
#include <stdio.h>
#include <stdlib.h>
struct node {
int key;
struct node *left, *right;
};
// A utility 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;
}
// A utility 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);
}
}
// A utility 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 (unchanged) node pointer
return node;
}
// Driver Code
int main()
{
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
struct node* root = NULL;
root = insert(root, 50);
insert(root, 30);
insert(root, 20);
insert(root, 40);
insert(root, 70);
insert(root, 60);
insert(root, 80);
// Print inorder traversal of the BST
inorder(root);
return 0;
}
C++14
// C++ program to demonstrate insertion
// in a BST recursively
#include <bits/stdc++.h>
using namespace std;
class BST {
int data;
BST *left, *right;
public:
// Default constructor.
BST();
// Parameterized constructor.
BST(int);
// Insert function.
BST* Insert(BST*, int);
// Inorder traversal.
void Inorder(BST*);
};
// Default Constructor definition.
BST::BST()
: data(0), left(NULL), right(NULL)
{
}
// Parameterized Constructor definition.
BST::BST(int value)
{
data = value;
left = right = NULL;
}
// Insert function definition.
BST* BST::Insert(BST* root, int value)
{
if (!root) {
// Insert the first node, if root is NULL.
return new BST(value);
}
// Insert data.
if (value > root->data) {
// Insert right node data, if the 'value'
// to be inserted is greater than 'root' node data.
// Process right nodes.
root->right = Insert(root->right, value);
}
else if (value < root->data) {
// Insert left node data, if the 'value'
// to be inserted is smaller than 'root' node data.
// Process left nodes.
root->left = Insert(root->left, value);
}
// Return 'root' node, after insertion.
return root;
}
// Inorder traversal function.
// This gives data in sorted order.
void BST::Inorder(BST* root)
{
if (!root) {
return;
}
Inorder(root->left);
cout << root->data << " ";
Inorder(root->right);
}
// Driver code
int main()
{
BST b, *root = NULL;
root = b.Insert(root, 50);
b.Insert(root, 30);
b.Insert(root, 20);
b.Insert(root, 40);
b.Insert(root, 70);
b.Insert(root, 60);
b.Insert(root, 80);
b.Inorder(root);
return 0;
}
Java
// Java program to demonstrate
// insert operation in binary
// search tree
import java.io.*;
public class BinarySearchTree {
// Class containing left
// and right child of current node
// and key value
class Node {
int key;
Node left, right;
public Node(int item)
{
key = item;
left = right = null;
}
}
// Root of BST
Node root;
// Constructor
BinarySearchTree() { root = null; }
BinarySearchTree(int value) { root = new Node(value); }
// This method mainly calls insertRec()
void insert(int key) { root = insertRec(root, key); }
// A recursive function to
// insert a new key in BST
Node insertRec(Node root, int key)
{
// If the tree is empty,
// return a new node
if (root == null) {
root = new Node(key);
return root;
}
// Otherwise, recur down the tree
else if (key < root.key)
root.left = insertRec(root.left, key);
else if (key > root.key)
root.right = insertRec(root.right, key);
// Return the (unchanged) node pointer
return root;
}
// This method mainly calls InorderRec()
void inorder() { inorderRec(root); }
// A utility function to
// do inorder traversal of BST
void inorderRec(Node root)
{
if (root != null) {
inorderRec(root.left);
System.out.print(root.key + " ");
inorderRec(root.right);
}
}
// Driver Code
public static void main(String[] args)
{
BinarySearchTree tree = new BinarySearchTree();
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
tree.insert(50);
tree.insert(30);
tree.insert(20);
tree.insert(40);
tree.insert(70);
tree.insert(60);
tree.insert(80);
// Print inorder traversal of the BST
tree.inorder();
}
}
// This code is contributed by Ankur Narain Verma
Python3
# 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, end =" ")
inorder(root.right)
# Driver program to test the above functions
if __name__ == '__main__':
# 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 inorder traversal of the BST
inorder(r)
C#
// C# program to demonstrate
// insert operation in binary
// search tree
using System;
class BinarySearchTree {
// Class containing left and
// right child of current node
// and key value
public class Node {
public int key;
public Node left, right;
public Node(int item)
{
key = item;
left = right = null;
}
}
// Root of BST
Node root;
// Constructor
BinarySearchTree() { root = null; }
BinarySearchTree(int value) { root = new Node(value); }
// This method mainly calls insertRec()
void insert(int key) { root = insertRec(root, key); }
// A recursive function to insert
// a new key in BST
Node insertRec(Node root, int key)
{
// If the tree is empty,
// return a new node
if (root == null) {
root = new Node(key);
return root;
}
// Otherwise, recur down the tree
if (key < root.key)
root.left = insertRec(root.left, key);
else if (key > root.key)
root.right = insertRec(root.right, key);
// Return the (unchanged) node pointer
return root;
}
// This method mainly calls InorderRec()
void inorder() { inorderRec(root); }
// A utility function to
// do inorder traversal of BST
void inorderRec(Node root)
{
if (root != null) {
inorderRec(root.left);
Console.Write(root.key + " ");
inorderRec(root.right);
}
}
// Driver Code
public static void Main(String[] args)
{
BinarySearchTree tree = new BinarySearchTree();
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
tree.insert(50);
tree.insert(30);
tree.insert(20);
tree.insert(40);
tree.insert(70);
tree.insert(60);
tree.insert(80);
// Print inorder traversal of the BST
tree.inorder();
}
}
// This code is contributed by aashish1995
Javascript
<script>
// javascript program to demonstrate
// insert operation in binary
// search tree
/*
* Class containing left and right child of current node and key value
*/
class Node {
constructor(item) {
this.key = item;
this.left = this.right = null;
}
}
// Root of BST
var root = null;
// This method mainly calls insertRec()
function insert(key) {
root = insertRec(root, key);
}
// A recursive function to insert a new key in BST
function insertRec(root, key) {
// If the tree is empty, return a new node
if (root == null) {
root = new Node(key);
return root;
}
// Otherwise, recur down the tree
if (key < root.key)
root.left = insertRec(root.left, key);
else if (key > root.key)
root.right = insertRec(root.right, key);
// Return the (unchanged) node pointer
return root;
}
// This method mainly calls InorderRec()
function inorder() {
inorderRec(root);
}
// A utility function to
// do inorder traversal of BST
function inorderRec(root)
{
if (root != null) {
inorderRec(root.left);
document.write(root.key+"<br/>");
inorderRec(root.right);
}
}
// Driver Code
/* Let us create following BST
50
/ \
30 70
/ \ / \
20 40 60 80 */
insert(50);
insert(30);
insert(20);
insert(40);
insert(70);
insert(60);
insert(80);
// Print inorder traversal of the BST
inorder();
// This code is contributed by Rajput-Ji
</script>
Output
20 30 40 50 60 70 80
Time Complexity:
- The worst-case time complexity of 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 the root to the deepest leaf node. The height of a skewed tree may become n and the time complexity of insertion operation may become O(n).
Auxiliary Space: The auxiliary space complexity of insertion into a binary search tree is O(1)
Example no 2 – Implementation using Iterative approach:
Instead of using recursion, we can also implement the insertion operation iteratively using a while loop. Below is the implementation using a while loop.
C++
// C++ Code to insert node and to print inorder traversal
// using iteration
#include <bits/stdc++.h>
using namespace std;
// BST Node
class Node {
public:
int val;
Node* left;
Node* right;
Node(int val)
: val(val), left(NULL), right(NULL)
{
}
};
// Utility function to insert node in BST
void insert(Node*& root, int key)
{
Node* node = new Node(key);
if (!root) {
root = node;
return;
}
Node* prev = NULL;
Node* temp = root;
while (temp) {
if (temp->val > key) {
prev = temp;
temp = temp->left;
}
else if (temp->val < key) {
prev = temp;
temp = temp->right;
}
}
if (prev->val > key)
prev->left = node;
else
prev->right = node;
}
// Utility function to print inorder traversal
void inorder(Node* root)
{
Node* temp = root;
stack<Node*> st;
while (temp != NULL || !st.empty()) {
if (temp != NULL) {
st.push(temp);
temp = temp->left;
}
else {
temp = st.top();
st.pop();
cout << temp->val << " ";
temp = temp->right;
}
}
}
// Driver code
int main()
{
Node* root = NULL;
insert(root, 30);
insert(root, 50);
insert(root, 15);
insert(root, 20);
insert(root, 10);
insert(root, 40);
insert(root, 60);
// Function call to print the inorder traversal
inorder(root);
return 0;
}
Java
// Java code to implement the insertion
// in binary search tree
import java.io.*;
import java.util.*;
class GFG {
// Driver code
public static void main(String[] args)
{
BST tree = new BST();
tree.insert(30);
tree.insert(50);
tree.insert(15);
tree.insert(20);
tree.insert(10);
tree.insert(40);
tree.insert(60);
tree.inorder();
}
}
class Node {
Node left;
int val;
Node right;
Node(int val) { this.val = val; }
}
class BST {
Node root;
// Function to insert a key
public void insert(int key)
{
Node node = new Node(key);
if (root == null) {
root = node;
return;
}
Node prev = null;
Node temp = root;
while (temp != null) {
if (temp.val > key) {
prev = temp;
temp = temp.left;
}
else if (temp.val < key) {
prev = temp;
temp = temp.right;
}
}
if (prev.val > key)
prev.left = node;
else
prev.right = node;
}
// Function to print the inorder value
public void inorder()
{
Node temp = root;
Stack<Node> stack = new Stack<>();
while (temp != null || !stack.isEmpty()) {
if (temp != null) {
stack.add(temp);
temp = temp.left;
}
else {
temp = stack.pop();
System.out.print(temp.val + " ");
temp = temp.right;
}
}
}
}
Python3
# Python 3 code to implement the insertion
# operation iteratively
class GFG:
@staticmethod
def main(args):
tree = BST()
tree.insert(30)
tree.insert(50)
tree.insert(15)
tree.insert(20)
tree.insert(10)
tree.insert(40)
tree.insert(60)
tree.inorder()
class Node:
left = None
val = 0
right = None
def __init__(self, val):
self.val = val
class BST:
root = None
# Function to insert a key in the BST
def insert(self, key):
node = Node(key)
if (self.root == None):
self.root = node
return
prev = None
temp = self.root
while (temp != None):
if (temp.val > key):
prev = temp
temp = temp.left
elif(temp.val < key):
prev = temp
temp = temp.right
if (prev.val > key):
prev.left = node
else:
prev.right = node
# Function to print the inorder traversal of BST
def inorder(self):
temp = self.root
stack = []
while (temp != None or not (len(stack) == 0)):
if (temp != None):
stack.append(temp)
temp = temp.left
else:
temp = stack.pop()
print(str(temp.val) + " ", end ="")
temp = temp.right
if __name__ == "__main__":
GFG.main([])
# This code is contributed by rastogik346.
C#
// Function to implement the insertion
// operation iteratively
using System;
using System.Collections.Generic;
public class GFG {
// Driver code
public static void Main(String[] args)
{
BST tree = new BST();
tree.insert(30);
tree.insert(50);
tree.insert(15);
tree.insert(20);
tree.insert(10);
tree.insert(40);
tree.insert(60);
// Function call to print the inorder traversal
tree.inorder();
}
}
public class Node {
public Node left;
public int val;
public Node right;
public Node(int val) { this.val = val; }
}
public class BST {
public Node root;
// Function to insert a new key in the BST
public void insert(int key)
{
Node node = new Node(key);
if (root == null) {
root = node;
return;
}
Node prev = null;
Node temp = root;
while (temp != null) {
if (temp.val > key) {
prev = temp;
temp = temp.left;
}
else if (temp.val < key) {
prev = temp;
temp = temp.right;
}
}
if (prev.val > key)
prev.left = node;
else
prev.right = node;
}
// Function to print the inorder traversal of BST
public void inorder()
{
Node temp = root;
Stack<Node> stack = new Stack<Node>();
while (temp != null || stack.Count != 0) {
if (temp != null) {
stack.Push(temp);
temp = temp.left;
}
else {
temp = stack.Pop();
Console.Write(temp.val + " ");
temp = temp.right;
}
}
}
}
// This code is contributed by Rajput-Ji
Javascript
// JavaScript code to implement the insertion
// in binary search tree
class Node {
constructor(val) {
this.left = null;
this.val = val;
this.right = null;
}
}
class BST {
constructor() {
this.root = null;
}
// Function to insert a key
insert(key) {
let node = new Node(key);
if (this.root == null) {
this.root = node;
return;
}
let prev = null;
let temp = this.root;
while (temp != null) {
if (temp.val > key) {
prev = temp;
temp = temp.left;
} else if (temp.val < key) {
prev = temp;
temp = temp.right;
}
}
if (prev.val > key) prev.left = node;
else prev.right = node;
}
// Function to print the inorder value
inorder() {
let temp = this.root;
let stack = [];
while (temp != null || stack.length > 0) {
if (temp != null) {
stack.push(temp);
temp = temp.left;
} else {
temp = stack.pop();
console.log(temp.val + " ");
temp = temp.right;
}
}
}
}
let tree = new BST();
tree.insert(30);
tree.insert(50);
tree.insert(15);
tree.insert(20);
tree.insert(10);
tree.insert(40);
tree.insert(60);
tree.inorder();
// this code is contributed by devendrasolunke
Output
10 15 20 30 40 50 60
The time complexity of inserting a node in a BST is O(log n), as we need to traverse down the tree to insert the node.
The Auxiliary space is O(1), as we do not use any extra space while inserting the node.
The time complexity of inorder traversal is O(n), as each node is visited once.
The Auxiliary space is O(n), as we use a stack to store nodes for recursion.
Example no 3 – Algorithmic Steps for Insertion Operation (With example)
The insertion operation in a BST can be explained in detail as follows:
- Initialize a pointer curr to the root node of the tree. If the tree is empty, create a new node with the given data and make it the root node.
- Compare the value of the new node with the value of the curr node.
- If the value of the new node is less than the value of the curr node, move the pointer curr to the left child of the curr node.
If the value of the new node is greater than the value of the curr node, move the pointer curr to the right child of the curr node.
Repeat step 2 until you reach an empty child node.- Create a new node with the given data at the location of the empty child node and connect it to its parent node.
- Repeat this process until the new node is inserted in its correct place in the BST.
Here is an example to illustrate the insertion operation:
Suppose, we want to insert the value 25 into the following BST:
C++
50 / \ 30 70 / \ / \ 20 40 60 80
- The pointer curr is initialized to the root node with value 50.
- The value of the new node 25 is less than the value of the curr node 50, so the pointer curr moves to the left child node with value 30.
- The value of the new node 25 is less than the value of the curr node 30, so the pointer curr moves to the left child node with value 20.
- The curr node has no left child node, so we create a new node with value 25 and connect it to its parent node 20.
The BST after inserting the value 25 will look like this:
C
50 / \ 30 70 / \ / 20 40 60 80 / 25Note that the insertion operation in a BST takes O(log n) time on average. However, in the worst case scenario, when the tree is skewed, it takes O(n) time. Therefore, it is important to maintain a balanced tree to ensure good performance.
Related Links:
- Binary Search Tree Delete Operation
- Quiz on Binary Search Tree
- Coding practice on BST
- All Articles on BST
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