Binary Search Tree (BST) Traversals – Inorder, Preorder, Post Order
Last Updated :
23 Jul, 2025
Given a Binary Search Tree, The task is to print the elements in inorder, preorder, and postorder traversal of the Binary Search Tree.
Input:
A Binary Search TreeOutput:
Inorder Traversal: 10 20 30 100 150 200 300
Preorder Traversal: 100 20 10 30 200 150 300
Postorder Traversal: 10 30 20 150 300 200 100
Input:
Binary Search TreeOutput:
Inorder Traversal: 8 12 20 22 25 30 40
Preorder Traversal: 22 12 8 20 30 25 40
Postorder Traversal: 8 20 12 25 40 30 22
Below is the idea to solve the problem:
At first traverse left subtree then visit the root and then traverse the right subtree.
Follow the below steps to implement the idea:
- Traverse left subtree
- Visit the root and print the data.
- Traverse the right subtree
The inorder traversal of the BST gives the values of the nodes in sorted order. To get the decreasing order visit the right, root, and left subtree.
Below is the implementation of the inorder traversal.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Class describing a node of tree
class Node {
public:
int data;
Node* left;
Node* right;
Node(int v)
{
this->data = v;
this->left = this->right = NULL;
}
};
// Inorder Traversal
void printInorder(Node* node)
{
if (node == NULL)
return;
// Traverse left subtree
printInorder(node->left);
// Visit node
cout << node->data << " ";
// Traverse right subtree
printInorder(node->right);
}
// Driver code
int main()
{
// Build the tree
Node* root = new Node(100);
root->left = new Node(20);
root->right = new Node(200);
root->left->left = new Node(10);
root->left->right = new Node(30);
root->right->left = new Node(150);
root->right->right = new Node(300);
// Function call
cout << "Inorder Traversal: ";
printInorder(root);
return 0;
}
// Java code to implement the approach
import java.io.*;
// Class describing a node of tree
class Node {
int data;
Node left;
Node right;
Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
class GFG {
// Inorder Traversal
public static void printInorder(Node node)
{
if (node == null)
return;
// Traverse left subtree
printInorder(node.left);
// Visit node
System.out.print(node.data + " ");
// Traverse right subtree
printInorder(node.right);
}
// Driver Code
public static void main(String[] args)
{
// Build the tree
Node root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
System.out.print("Inorder Traversal: ");
printInorder(root);
}
}
// This code is contributed by Rohit Pradhan
# Python3 code to implement the approach
# Class describing a node of tree
class Node:
def __init__(self, v):
self.left = None
self.right = None
self.data = v
# Inorder Traversal
def printInorder(root):
if root:
# Traverse left subtree
printInorder(root.left)
# Visit node
print(root.data,end=" ")
# Traverse right subtree
printInorder(root.right)
# Driver code
if __name__ == "__main__":
# Build the tree
root = Node(100)
root.left = Node(20)
root.right = Node(200)
root.left.left = Node(10)
root.left.right = Node(30)
root.right.left = Node(150)
root.right.right = Node(300)
# Function call
print("Inorder Traversal:",end=" ")
printInorder(root)
# This code is contributed by ajaymakvana.
// Include namespace system
using System;
// Class describing a node of tree
public class Node
{
public int data;
public Node left;
public Node right;
public Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
public class GFG
{
// Inorder Traversal
public static void printInorder(Node node)
{
if (node == null)
{
return;
}
// Traverse left subtree
GFG.printInorder(node.left);
// Visit node
Console.Write(node.data.ToString() + " ");
// Traverse right subtree
GFG.printInorder(node.right);
}
// Driver Code
public static void Main(String[] args)
{
// Build the tree
var root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
Console.Write("Inorder Traversal: ");
GFG.printInorder(root);
}
}
// JavaScript code to implement the approach
class Node {
constructor(v) {
this.left = null;
this.right = null;
this.data = v;
}
}
// Inorder Traversal
function printInorder(root)
{
if (root)
{
// Traverse left subtree
printInorder(root.left);
// Visit node
console.log(root.data);
// Traverse right subtree
printInorder(root.right);
}
}
// Driver code
if (true)
{
// Build the tree
let root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
console.log("Inorder Traversal:");
printInorder(root);
}
// This code is contributed by akashish__
OutputInorder Traversal: 10 20 30 100 150 200 300
Time complexity: O(N), Where N is the number of nodes.
Auxiliary Space: O(h), Where h is the height of tree
Below is the idea to solve the problem:
At first visit the root then traverse left subtree and then traverse the right subtree.
Follow the below steps to implement the idea:
- Visit the root and print the data.
- Traverse left subtree
- Traverse the right subtree
Below is the implementation of the preorder traversal.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Class describing a node of tree
class Node {
public:
int data;
Node* left;
Node* right;
Node(int v)
{
this->data = v;
this->left = this->right = NULL;
}
};
// Preorder Traversal
void printPreOrder(Node* node)
{
if (node == NULL)
return;
// Visit Node
cout << node->data << " ";
// Traverse left subtree
printPreOrder(node->left);
// Traverse right subtree
printPreOrder(node->right);
}
// Driver code
int main()
{
// Build the tree
Node* root = new Node(100);
root->left = new Node(20);
root->right = new Node(200);
root->left->left = new Node(10);
root->left->right = new Node(30);
root->right->left = new Node(150);
root->right->right = new Node(300);
// Function call
cout << "Preorder Traversal: ";
printPreOrder(root);
return 0;
}
// Java code to implement the approach
import java.io.*;
// Class describing a node of tree
class Node {
int data;
Node left;
Node right;
Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
class GFG {
// Preorder Traversal
public static void printPreorder(Node node)
{
if (node == null)
return;
// Visit node
System.out.print(node.data + " ");
// Traverse left subtree
printPreorder(node.left);
// Traverse right subtree
printPreorder(node.right);
}
public static void main(String[] args)
{
// Build the tree
Node root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
System.out.print("Preorder Traversal: ");
printPreorder(root);
}
}
// This code is contributed by lokeshmvs21.
class Node:
def __init__(self, v):
self.data = v
self.left = None
self.right = None
# Preorder Traversal
def printPreOrder(node):
if node is None:
return
# Visit Node
print(node.data, end = " ")
# Traverse left subtree
printPreOrder(node.left)
# Traverse right subtree
printPreOrder(node.right)
# Driver code
if __name__ == "__main__":
# Build the tree
root = Node(100)
root.left = Node(20)
root.right = Node(200)
root.left.left = Node(10)
root.left.right = Node(30)
root.right.left = Node(150)
root.right.right = Node(300)
# Function call
print("Preorder Traversal: ", end = "")
printPreOrder(root)
// Include namespace system
using System;
// Class describing a node of tree
public class Node
{
public int data;
public Node left;
public Node right;
public Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
public class GFG
{
// Preorder Traversal
public static void printPreorder(Node node)
{
if (node == null)
{
return;
}
// Visit node
Console.Write(node.data.ToString() + " ");
// Traverse left subtree
GFG.printPreorder(node.left);
// Traverse right subtree
GFG.printPreorder(node.right);
}
public static void Main(String[] args)
{
// Build the tree
var root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
Console.Write("Preorder Traversal: ");
GFG.printPreorder(root);
}
}
class Node {
constructor(v) {
this.data = v;
this.left = this.right = null;
}
}
function printPreOrder(node) {
if (node == null) return;
console.log(node.data + " ");
printPreOrder(node.left);
printPreOrder(node.right);
}
// Build the tree
let root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
console.log("Preorder Traversal: ");
printPreOrder(root);
// This code is contributed by akashish__
OutputPreorder Traversal: 100 20 10 30 200 150 300
Time complexity: O(N), Where N is the number of nodes.
Auxiliary Space: O(H), Where H is the height of the tree
Below is the idea to solve the problem:
At first traverse left subtree then traverse the right subtree and then visit the root.
Follow the below steps to implement the idea:
- Traverse left subtree
- Traverse the right subtree
- Visit the root and print the data.
Below is the implementation of the postorder traversal:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Class to define structure of a node
class Node {
public:
int data;
Node* left;
Node* right;
Node(int v)
{
this->data = v;
this->left = this->right = NULL;
}
};
// PostOrder Traversal
void printPostOrder(Node* node)
{
if (node == NULL)
return;
// Traverse left subtree
printPostOrder(node->left);
// Traverse right subtree
printPostOrder(node->right);
// Visit node
cout << node->data << " ";
}
// Driver code
int main()
{
Node* root = new Node(100);
root->left = new Node(20);
root->right = new Node(200);
root->left->left = new Node(10);
root->left->right = new Node(30);
root->right->left = new Node(150);
root->right->right = new Node(300);
// Function call
cout << "PostOrder Traversal: ";
printPostOrder(root);
cout << "\n";
return 0;
}
// Java code to implement the approach
import java.io.*;
// Class describing a node of tree
class GFG {
static class Node {
int data;
Node left;
Node right;
Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
// Preorder Traversal
public static void printPostOrder(Node node)
{
if (node == null)
return;
// Traverse left subtree
printPostOrder(node.left);
// Traverse right subtree
printPostOrder(node.right);
// Visit node
System.out.print(node.data + " ");
}
public static void main(String[] args)
{
// Build the tree
Node root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
System.out.print("PostOrder Traversal: ");
printPostOrder(root);
}
}
class Node:
def __init__(self, v):
self.data = v
self.left = None
self.right = None
# Preorder Traversal
def printPostOrder(node):
if node is None:
return
# Traverse left subtree
printPostOrder(node.left)
# Traverse right subtree
printPostOrder(node.right)
# Visit Node
print(node.data, end = " ")
# Driver code
if __name__ == "__main__":
# Build the tree
root = Node(100)
root.left = Node(20)
root.right = Node(200)
root.left.left = Node(10)
root.left.right = Node(30)
root.right.left = Node(150)
root.right.right = Node(300)
# Function call
print("Postorder Traversal: ", end = "")
printPostOrder(root)
// Include namespace system
using System;
// Class describing a node of tree
public class Node
{
public int data;
public Node left;
public Node right;
public Node(int v)
{
this.data = v;
this.left = this.right = null;
}
}
public class GFG
{
// Preorder Traversal
public static void printPostOrder(Node node)
{
if (node == null)
{
return;
}
// Traverse left subtree
GFG.printPostOrder(node.left);
// Traverse right subtree
GFG.printPostOrder(node.right);
// Visit node
Console.Write(node.data.ToString() + " ");
}
public static void Main(String[] args)
{
// Build the tree
var root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
Console.Write("PostOrder Traversal: ");
GFG.printPostOrder(root);
}
}
class Node {
constructor(v) {
this.data = v;
this.left = null;
this.right = null;
}
}
// Preorder Traversal
function printPostOrder(node) {
if (node === null) {
return;
}
// Traverse left subtree
printPostOrder(node.left);
// Traverse right subtree
printPostOrder(node.right);
// Visit Node
console.log(node.data, end = " ");
}
// Driver code
// Build the tree
let root = new Node(100);
root.left = new Node(20);
root.right = new Node(200);
root.left.left = new Node(10);
root.left.right = new Node(30);
root.right.left = new Node(150);
root.right.right = new Node(300);
// Function call
console.log("Postorder Traversal: ", end = "");
printPostOrder(root);
// This code is contributed by akashish__
OutputPostOrder Traversal: 10 30 20 150 300 200 100
Time complexity: O(N), Where N is the number of nodes.
Auxiliary Space: O(H), Where H is the height of the tree
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