检查一个数组是否代表二叉查找树的整数
原文:https://www . geesforgeks . org/check-array-representatives-inoder-binary-search-tree-not/
给定一组 N 元素。任务是检查它是否是任意二叉查找树的有序遍历。打印“是”，如果它是任何其他二叉查找树的有序遍历打印“否”。 例:
Input : arr[] = { 19, 23, 25, 30, 45 }
Output : Yes
Input : arr[] = { 19, 23, 30, 25, 45 }
Output : No
这个想法是利用二叉查找树的有序遍历是有序的这一事实。所以，只要检查给定的数组是否排序。
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 << "Yesn";
else
cout << "Non";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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("Non");
}
}
}
//This code is contributed by 29AjayKumar
Python 3
# 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("Non");
}
}
}
// This code is contributed by Rajput-Ji
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<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("Non");
}
// This code is contributed by Rajput-Ji
</script>
输出:
Yes
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