Given an array of N element. The task is to check if it is Inorder traversal of any Binary Search Tree or not. Print “Yes” if it is Inorder traversal of any Binary Search Tree else print “No”.
Examples:
Input : arr[] = { 19, 23, 25, 30, 45 } Output : Yes Input : arr[] = { 19, 23, 30, 25, 45 } Output : No
The idea is to use the fact that the inorder traversal of Binary Search Tree is sorted. So, just check if given array is sorted or not.
Implementation:
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( 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 ;
} // Driver code int main()
{ int arr[] = { 19, 23, 25, 30, 45 };
int n = sizeof (arr)/ sizeof (arr[0]);
if (isInorder(arr, n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
} |
Java
// Java program to check if a given array is sorted // or not. class GFG {
// Function that returns true if array is Inorder // traversal of any Binary Search Tree or not. static boolean 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 ;
}
// Drivers code public static void main(String[] args) {
int arr[] = { 19 , 23 , 25 , 30 , 45 };
int n = arr.length;
if (isInorder(arr, n)) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
} //This code is contributed by 29AjayKumar |
Python3
# Python 3 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):
# Array has one or no element
if (n = = 0 or n = = 1 ):
return True
for i in range ( 1 , n, 1 ):
# Unsorted pair found
if (arr[i - 1 ] > arr[i]):
return False
# No unsorted pair found
return True
# Driver code if __name__ = = '__main__' :
arr = [ 19 , 23 , 25 , 30 , 45 ]
n = len (arr)
if (isInorder(arr, n)):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed by # Sahil_Shelangia |
C#
// C# program to check if a given // array is sorted or not. using System;
class GFG
{ // Function that returns true if // array is Inorder traversal of // any Binary Search Tree or not. static bool isInorder( int [] arr, int n)
{ // Array has one or no element
if (n == 0 || n == 1)
{
return true ;
}
// Unsorted pair found
for ( int i = 1; i < n; i++)
{
if (arr[i - 1] > arr[i])
{
return false ;
}
}
// No unsorted pair found
return true ;
} // Driver code public static void Main()
{ int []arr = {19, 23, 25, 30, 45};
int n = arr.Length;
if (isInorder(arr, n))
{
Console.Write( "Yes" );
}
else
{
Console.Write( "No" );
}
} } // This code is contributed by Rajput-Ji |
PHP
<?php // PHP 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 , $n )
{ // Array has one or no element
if ( $n == 0 || $n == 1)
return true;
for ( $i = 1; $i < $n ; $i ++)
// Unsorted pair found
if ( $arr [ $i - 1] > $arr [ $i ])
return false;
// No unsorted pair found
return true;
} // Driver code $arr = array (19, 23, 25, 30, 45);
$n = sizeof( $arr );
if (isInorder( $arr , $n ))
echo "Yes" ;
else echo "No" ;
// This code is contributed // by Akanksha Rai ?> |
Javascript
<script> // 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 , n)
{
// Array has one or no element
if (n == 0 || n == 1) {
return true ;
}
for (i = 1; i < n; i++) // Unsorted pair found
{
if (arr[i - 1] > arr[i]) {
return false ;
}
}
// No unsorted pair found
return true ;
}
// Drivers code
var arr = [ 19, 23, 25, 30, 45 ];
var n = arr.length;
if (isInorder(arr, n)) {
document.write( "Yes" );
} else {
document.write( "No" );
}
// This code is contributed by Rajput-Ji </script> |
Output
Yes
Time complexity: O(n) where n is the size of array
Auxiliary Space: O(1)
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