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++
// CPP code to print level order
// traversal in sorted order
#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)
{
}
};
// Function to print sorted
// level order traversal
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);
}
}
}
// Driver code
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 code to print level order
// traversal in sorted order
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);
}
// Function to print sorted
// level order traversal
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);
}
}
}
// Driver Code
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);
}
}
// This code is contributed by SHUBHAMSINGH10
# Python3 program to print level order
# traversal in sorted order
# Helper function that allocates a new
# node with the given data and None
# left and right pointers.
class newNode:
# Construct to create a new node
def __init__(self, key):
self.data = key
self.left = None
self.right = None
# Function to print sorted
# level order traversal
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)
# Driver Code
if __name__ == '__main__':
"""
Let us create Binary Tree shown
in above example """
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)
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)
// C# code to print level order
// traversal in sorted order
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);
}
// Function to print sorted
// level order traversal
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);
}
}
}
// Driver Code
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);
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript code to print level order
// traversal in sorted order
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 to print sorted
// level order traversal
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);
}
}
}
// Driver Code
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);
// This code is contributed by rag2127
</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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