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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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`
1. Start traversing the array with level initialized as 0.
2. 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 ` `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); ` `    ``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 `

Output:

```5
4 6
1 2 9
```

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