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 ordervoid 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 codeint 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 approachimport java.util.Arrays;class GFG{// Function to print all the levels// of the given tree in sorted orderstatic 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 codepublic 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 approachusing System;using System.Linq; class GFG{// Function to print all the levels// of the given tree in sorted orderstatic 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 codepublic 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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