Check if an array represents Inorder of Binary Search tree or not Last Updated : 27 Dec, 2024 Summarize Comments Improve Suggest changes Share Like Article Like Report Try it on GfG Practice Given an array of n elements. The task is to check if it is an Inorder traversal of any Binary Search Tree or not.Examples: Input: arr[] = { 19, 23, 25, 30, 45 }Output: trueExplanation: As the array is sorted in non-decreasing order, it is an Inorder traversal of any Binary Search Tree. Input : arr[] = { 19, 23, 30, 25, 45 }Output : falseExplanation: As the array is not sorted in non-decreasing order, it is not an Inorder traversal of any Binary Search Tree.Approach:The idea is to use the fact that the in-order traversal of Binary Search Tree is sorted. So, just check if given array is sorted or not. Below is the implementation of the above approach: C++ // C++ program to check if a given array is sorted // or not. #include<bits/stdc++.h> using namespace std; // Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. bool isInorder(vector<int> &arr) { int n = arr.size(); // Array has one or no element if (n == 0 || n == 1) return true; for (int i = 1; i < n; i++) // Unsorted pair found if (arr[i-1] > arr[i]) return false; // No unsorted pair found return true; } int main() { vector<int> arr = { 19, 23, 25, 30, 45 }; if (isInorder(arr)) cout << "true"; else cout << "false"; return 0; } C // C program to check if a given array is sorted // or not. #include <stdio.h> #include <stdbool.h> // Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. bool isInorder(int arr[], int n) { // Array has one or no element if (n == 0 || n == 1) return true; for (int i = 1; i < n; i++) { // Unsorted pair found if (arr[i-1] > arr[i]) return false; } // No unsorted pair found return true; } int main() { int arr[] = { 19, 23, 25, 30, 45 }; int n = sizeof(arr) / sizeof(arr[0]); if (isInorder(arr, n)) printf("true"); else printf("false"); return 0; } Java // Java program to check if a given array is sorted // or not. import java.util.ArrayList; class GfG { // Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. static boolean isInorder(ArrayList<Integer> arr) { int n = arr.size(); // Array has one or no element if (n == 0 || n == 1) return true; for (int i = 1; i < n; i++) { // Unsorted pair found if (arr.get(i - 1) > arr.get(i)) return false; } // No unsorted pair found return true; } public static void main(String[] args) { ArrayList<Integer> arr = new ArrayList<>(); arr.add(19); arr.add(23); arr.add(25); arr.add(30); arr.add(45); if (isInorder(arr)) System.out.println("true"); else System.out.println("false"); } } Python # Python program to check if a given array is sorted # or not. # Function that returns true if array is Inorder # traversal of any Binary Search Tree or not. def isInorder(arr): n = len(arr) # Array has one or no element if n == 0 or n == 1: return True for i in range(1, n): # Unsorted pair found if arr[i-1] > arr[i]: return False # No unsorted pair found return True if __name__ == "__main__": arr = [19, 23, 25, 30, 45] if isInorder(arr): print("true") else: print("false") C# // C# program to check if a given array is sorted // or not. using System; using System.Collections.Generic; class GfG { // Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. static bool isInorder(List<int> arr) { int n = arr.Count; // Array has one or no element if (n == 0 || n == 1) return true; for (int i = 1; i < n; i++) { // Unsorted pair found if (arr[i-1] > arr[i]) return false; } // No unsorted pair found return true; } static void Main() { List<int> arr = new List<int> { 19, 23, 25, 30, 45 }; if (isInorder(arr)) Console.WriteLine("true"); else Console.WriteLine("false"); } } JavaScript // JavaScript program to check if a given array is sorted // or not. // Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. function isInorder(arr) { let n = arr.length; // Array has one or no element if (n === 0 || n === 1) return true; for (let i = 1; i < n; i++) { // Unsorted pair found if (arr[i-1] > arr[i]) return false; } // No unsorted pair found return true; } const arr = [19, 23, 25, 30, 45]; if (isInorder(arr)) console.log("true"); else console.log("false"); OutputtrueTime complexity: O(n) where n is the size of array.Auxiliary Space: O(1) Inorder Traversal and BST | DSA Problem Comment More infoAdvertise with us Next Article Introduction to Binary Search Tree K kartik Improve Article Tags : Tree Binary Search Tree DSA Practice Tags : Binary Search TreeTree Similar Reads Binary Search Tree A Binary Search Tree (BST) is a type of binary tree data structure in which each node contains a unique key and satisfies a specific ordering property:All nodes in the left subtree of a node contain values strictly less than the node’s value. 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The height is typically maintained in order of logN so that all operations take O(logN) time on average 4 min read Like