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 traversalvoid 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 codeint 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
// Java code to print level order // traversal in sorted orderimport 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 traversalstatic 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 Codepublic 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
# 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 traversaldef 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#
// C# code to print level order // traversal in sorted orderusing 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 traversalstatic 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 Codepublic 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 |
Javascript
<script>// Javascript code to print level order // traversal in sorted ordervar 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 traversalfunction 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 Codelet 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> |
Output:
7 5 6 1 2 3 4
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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