Print Binary Tree levels in sorted order | Set 3 (Tree given as array)

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Given a Complete Binary Tree as an array, the task is to print all of its levels in sorted order.
Examples:

Input: arr[] = {7, 6, 5, 4, 3, 2, 1}
The given tree looks like
            7
          /   \
        6      5
       / \    / \
      4  3   2   1
Output:
7
5 6
1 2 3 4

Input: arr[] = {5, 6, 4, 9, 2, 1}
The given tree looks like 
            5
          /   \
        6      4
       / \    /    
      9   2  1
Output: 
5
4 6
1 2 9

Approach: A similar problem is discussed here
As the given tree is a Complete Binary Tree:

No. of nodes at a level l will be 2^l where l ? 0

Start traversing the array with level initialized as 0.
Sort the elements which are the part of the current level and print the elements.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print all the levels
// of the given tree in sorted order
void printSortedLevels(int arr[], int n)
{
 
    // Initialize level with 0
    int level = 0;
    for (int i = 0; i < n; level++) {
 
        // Number of nodes at current level
        int cnt = (int)pow(2, level);
 
        // Indexing of array starts from 0
        // so subtract no. of nodes by 1
        cnt -= 1;
 
        // Index of the last node in the current level
        int j = min(i + cnt, n - 1);
 
        // Sort the nodes of the current level
        sort(arr + i, arr + j + 1);
 
        // Print the sorted nodes
        while (i <= j) {
            cout << arr[i] << " ";
            i++;
        }
        cout << endl;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 4, 9, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printSortedLevels(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
 
class GFG
{
 
// Function to print all the levels
// of the given tree in sorted order
static void printSortedLevels(int arr[], int n)
{
 
    // Initialize level with 0
    int level = 0;
    for (int i = 0; i < n; level++) 
    {
 
        // Number of nodes at current level
        int cnt = (int)Math.pow(2, level);
 
        // Indexing of array starts from 0
        // so subtract no. of nodes by 1
        cnt -= 1;
 
        // Index of the last node in the current level
        int j = Math.min(i + cnt, n - 1);
 
        // Sort the nodes of the current level
        Arrays.sort(arr, i, j+1);
 
        // Print the sorted nodes
        while (i <= j) 
        {
            System.out.print(arr[i] + " ");
            i++;
        }
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 6, 4, 9, 2, 1 };
    int n = arr.length;
    printSortedLevels(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
from math import pow
 
# Function to print all the levels 
# of the given tree in sorted order 
def printSortedLevels(arr, n):
     
    # Initialize level with 0
    level = 0
    i = 0
    while(i < n):
         
        # Number of nodes at current level
        cnt = int(pow(2, level))
 
        # Indexing of array starts from 0
        # so subtract no. of nodes by 1
        cnt -= 1
 
        # Index of the last node in the current level
        j = min(i + cnt, n - 1)
         
        # Sort the nodes of the current level
        arr = arr[:i] + sorted(arr[i:j + 1]) + \
                               arr[j + 1:]
 
        # Print the sorted nodes
        while (i <= j):
            print(arr[i], end = " ")
            i += 1
        print()
        level += 1
 
# Driver code 
arr = [ 5, 6, 4, 9, 2, 1]
n = len(arr) 
printSortedLevels(arr, n) 
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;
using System.Linq;                 
     
class GFG
{
 
// Function to print all the levels
// of the given tree in sorted order
static void printSortedLevels(int []arr, int n)
{
 
    // Initialize level with 0
    int level = 0;
    for (int i = 0; i < n; level++) 
    {
 
        // Number of nodes at current level
        int cnt = (int)Math.Pow(2, level);
 
        // Indexing of array starts from 0
        // so subtract no. of nodes by 1
        cnt -= 1;
 
        // Index of the last node in the current level
        int j = Math.Min(i + cnt, n - 1);
 
        // Sort the nodes of the current level
        Array.Sort(arr, i, j + 1 - i);
 
        // Print the sorted nodes
        while (i <= j) 
        {
            Console.Write(arr[i] + " ");
            i++;
        }
        Console.WriteLine();
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 5, 6, 4, 9, 2, 1 };
    int n = arr.Length;
    printSortedLevels(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
    // JavaScript implementation of the approach
     
    // Function to sort the elements of the array
    // from index a to index b
    function partSort(arr, N, a, b)
    {
        // Variables to store start and end of the index range
        let l = Math.min(a, b);
        let r = Math.max(a, b);
   
        // Temporary array
        let temp = new Array(r - l + 1);
        temp.fill(0);
        let j = 0;
        for (let i = l; i <= r; i++) {
            temp[j] = arr[i];
            j++;
        }
   
        // Sort the temporary array
        temp.sort(function(a, b){return a - b});
   
        // Modifying original array with temporary array elements
        j = 0;
        for (let i = l; i <= r; i++) {
            arr[i] = temp[j];
            j++;
        }
    }
     
    // Function to print all the levels
    // of the given tree in sorted order
    function printSortedLevels(arr, n)
    {
 
        // Initialize level with 0
        let level = 0;
        for (let i = 0; i < n; level++) 
        {
 
            // Number of nodes at current level
            let cnt = Math.pow(2, level);
 
            // Indexing of array starts from 0
            // so subtract no. of nodes by 1
            cnt -= 1;
 
            // Index of the last node in the current level
            let j = Math.min(i + cnt, n - 1);
 
            // Sort the nodes of the current level
            partSort(arr, n, i, j + 1);
 
            // Print the sorted nodes
            while (i <= j) 
            {
                document.write(arr[i] + " ");
                i++;
            }
            document.write("</br>");
        }
    }
     
    let arr = [ 5, 6, 4, 9, 2, 1 ];
    let n = arr.length;
    printSortedLevels(arr, n);
     
</script>

Output:

5 
4 6 
1 2 9

Time complexity: O(nlogn) where n is no of nodes in binary tree

Auxiliary space: O(1) as it is using constant space

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