Print Binary Tree levels in sorted order | Set 3 (Tree given as array)
Last Updated :
11 Jul, 2025
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 2l 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 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 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# 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
<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>
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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