# How to insert a Node in a Binary Search Tree

Recommended to visit before going through this article:
What is a Binary Search Tree? How to represent it ? And, related terminologies…

Inserting a node in a given Binary Search Tree is a process to add a new node; let’s say if node A has to be inserted then you got to follow below steps –

STEP 1: If there is no node in a given BST then insert node A as its Root Node.

STEP 2: Find the Node in a given Binary Search Tree where we need to insert A ( as either left or a right child ). To find so, just compare the value of node A with the root node’s value:

if node A has a value greater than the root node’s value – traverse down the root node in its right node and Go to Step 2 by considering this right node as the root node (Note: if there is no right node straight go to Step 3)

if node A has a value lesser than the root node’s value – traverse down the root node in its left node and Go to Step 2 by considering this left node as the root node (Note: if there is no left node straight go to Step 3)

STEP 3: Once the Node is found using step 1:

if value of node A is greater than this Node’s value – insert A as the right child of this node
if value of node A is lesser than this Node’s value – insert A as the left child of this node

### Above Algorithm can be implemented using two popular ways – Recursive and an Iterative way

BST,java

```
package org.gontuseries.bst;

public class BST {

private Node root;

public void insert(int value) {

root = insert(value, root);
}

private Node insert(int value, Node currentNode) {

if (currentNode == null) {
return createNewNode(value);
}

if (value > currentNode.getValue()) {
currentNode = insert(value, currentNode.getRightNode());
} else if (value < currentNode.getValue()) {
currentNode = insert(value, currentNode.getLeftNode());
} else {
System.out.println("Node with the same value already
exists in the BST");
}

return currentNode;
}

private Node createNewNode(int value) {

Node node = new Node(value);

return node;
}
}

```
```
package org.gontuseries.bst;

public class BST {

private Node root;

private void insert(int value) {

if (root == null)
root = createNewNode(value);
else {
Node currentNode = root;
while (true) {

if (value > currentNode.getValue()) {
if (currentNode.getRightNode() != null) {
currentNode = currentNode.getRightNode();
} else {
currentNode.setRightNode(createNewNode(value));
}
} else if (value < currentNode.getValue()) {
if (currentNode.getLeftNode() != null) {
currentNode = currentNode.getLeftNode();
} else {
currentNode.setLeftNode(createNewNode(value));
}
} else {
System.out.println("Node with the same value already
exists in the BST");
}
}
}
}

private Node createNewNode(int value) {

Node node = new Node(value);
return node;
}
}

```

Node.java

```
package org.gontuseries.bst;

public class Node {

Node(int value) {
this.value = value;
this.leftNode = null;
this.rightNode = null;
}

private int value;
private Node leftNode;
private Node rightNode;
// --- writing all getters and setters for all properties
//i.e for value, leftNode, rightNode
}
```

Time Complexity: The run time complexity of insert operation using Recursive way is: O(height of a Binary Search Tree) i.e O(h) [worst-case]

a) In case of a  skewed Binary Search Tree the height is equal to the number of nodes in it; so, it becomes O(n)[worst-case]

b) In case of a Binary Search Tree built using some Tree Balancing Techniques like AVL, RED Black etc the height is equal to log (number of nodes in it); so it becomes log(n) [worst-case]

where, ‘n’ is the number of nodes in a binary search tree.