Print Binary Tree levels in sorted order | Set 2 (Using set)
Last Updated :
23 Jan, 2023
Given a tree, print the level order traversal in sorted order.
Examples :
Input : 7
/ \
6 5
/ \ / \
4 3 2 1
Output :
7
5 6
1 2 3 4
Input : 7
/ \
16 1
/ \
4 13
Output :
7
1 16
4 13
We have discussed a priority queue based solution in below post.
Print Binary Tree levels in sorted order | Set 1 (Using Priority Queue)
In this post, a set (which is implemented using balanced binary search tree) based solution is discussed.
Approach :
- Start level order traversal of tree.
- Store all the nodes in a set(or any other similar data structures).
- Print elements of set.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node* left;
Node* right;
Node( int dat = 0)
: data(dat), left(nullptr),
right(nullptr)
{
}
};
void sorted_level_order(Node* root)
{
queue<Node*> q;
set< int > s;
q.push(root);
q.push(nullptr);
while (q.empty() == false ) {
Node* tmp = q.front();
q.pop();
if (tmp == nullptr) {
if (s.empty() == true )
break ;
for (set< int >::iterator it =
s.begin();it != s.end(); ++it)
cout << *it << " " ;
q.push(nullptr);
s.clear();
}
else {
s.insert(tmp->data);
if (tmp->left != nullptr)
q.push(tmp->left);
if (tmp->right != nullptr)
q.push(tmp->right);
}
}
}
int main()
{
Node* root = new Node(7);
root->left = new Node(6);
root->right = new Node(5);
root->left->left = new Node(4);
root->left->right = new Node(3);
root->right->left = new Node(2);
root->right->right = new Node(1);
sorted_level_order(root);
return 0;
}
|
Java
import java.util.*;
import java.util.HashSet;
class GFG
{
static class Node
{
int data;
Node left, right;
};
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
}
static void sorted_level_order(Node root)
{
Queue<Node> q = new LinkedList<>();
Set<Integer> s = new HashSet<Integer>();
q.add(root);
q.add( null );
while (!q.isEmpty())
{
Node tmp = q.peek();
q.remove();
if (tmp == null )
{
if (s.isEmpty())
break ;
Iterator value = s.iterator();
while (value.hasNext())
{
System.out.print(value.next() + " " );
}
q.add( null );
s.clear();
}
else
{
s.add(tmp.data);
if (tmp.left != null )
q.add(tmp.left);
if (tmp.right != null )
q.add(tmp.right);
}
}
}
public static void main(String[] args)
{
Node root = newNode( 7 );
root.left = newNode( 6 );
root.right = newNode( 5 );
root.left.left = newNode( 4 );
root.left.right = newNode( 3 );
root.right.left = newNode( 2 );
root.right.right = newNode( 1 );
sorted_level_order(root);
}
}
|
Python3
class newNode:
def __init__( self , key):
self .data = key
self .left = None
self .right = None
def sorted_level_order( root):
q = []
s = set ()
q.append(root)
q.append( None )
while ( len (q)):
tmp = q[ 0 ]
q.pop( 0 )
if (tmp = = None ):
if ( not len (s)):
break
for i in s:
print (i, end = " " )
q.append( None )
s = set ()
else :
s.add(tmp.data)
if (tmp.left ! = None ):
q.append(tmp.left)
if (tmp.right ! = None ):
q.append(tmp.right)
if __name__ = = '__main__' :
root = newNode( 7 )
root.left = newNode( 6 )
root.right = newNode( 5 )
root.left.left = newNode( 4 )
root.left.right = newNode( 3 )
root.right.left = newNode( 2 )
root.right.right = newNode( 1 )
sorted_level_order(root)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
public class Node
{
public int data;
public Node left, right;
};
static Node newNode( int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
}
static void sorted_level_order(Node root)
{
Queue<Node> q = new Queue<Node>();
SortedSet< int > s = new SortedSet< int >();
q.Enqueue(root);
q.Enqueue( null );
while (q.Count != 0)
{
Node tmp = q.Peek();
q.Dequeue();
if (tmp == null )
{
if (s.Count == 0)
break ;
foreach ( int v in s)
{
Console.Write(v + " " );
}
q.Enqueue( null );
s.Clear();
}
else
{
s.Add(tmp.data);
if (tmp.left != null )
q.Enqueue(tmp.left);
if (tmp.right != null )
q.Enqueue(tmp.right);
}
}
}
public static void Main(String[] args)
{
Node root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
sorted_level_order(root);
}
}
|
Javascript
<script>
var SortedSet = require( "collections/sorted-set" );
class Node
{
constructor()
{
this .data=0;
this .left= this .right= null ;
}
}
function newNode(data)
{
let node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
}
function sorted_level_order(root)
{
let q = [];
let s = new SortedSet();
q.push(root);
q.push( null );
while (q.length!=0)
{
let tmp = q.shift();
if (tmp == null )
{
if (s.size==0)
break ;
for (let i of s.values())
{
document.write(i+ " " );
}
q.push( null );
s.clear();
}
else
{
s.add(tmp.data);
if (tmp.left != null )
q.push(tmp.left);
if (tmp.right != null )
q.push(tmp.right);
}
}
}
let root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
sorted_level_order(root);
</script>
|
Time Complexity: O(n*log(n)) where n is the number of nodes in the binary tree.
Auxiliary Space: O(n) where n is the number of nodes in the binary tree.
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