Given a binary tree with N nodes and an integer K, the task is to print nodes of the Kth level of a binary tree without duplicates.
Examples:
Input:
60 --- Level 0
/ \
50 30 --- Level 1
/ \ /
80 10 40 --- Level 2
K = 1
Output: 30 50
Input:
50 --- Level 0
/ \
60 70 --- Level 1
/ \ / \
90 40 40 20 --- Level 2
K = 2
Output: 20 40 90
Approach: The idea is to traverse the Binary Tree using the Level Order Traversal with the help of queue and if the Level of the Traversal is K then store all the Nodes of that Level in a Set such that there are no duplicate nodes at that level.
Algorithm:
- Initialize an Empty Queue to store the nodes at a level.
- Enqueue the Root node of the Binary Tree in the queue.
- Initialize the Level as 0, as the first level of the tree is supposed to be 0 here.
- Initialize the flag as 0 to check Kth level is reached or not.
- Iterate using a while loop until the queue is not empty.
- Find the size of the queue and store in a variable size to visit only the nodes of a current level.
- Iterate with another while loop until the size variable is not 0
- Deque a node from the queue and Enqueue its Left and right childs in the Queue.
- If the current level is equal to the K, then add the data of the node into the set and also set the flag.
- If flag is set then break the loop to not visit further levels, otherwise increment the current level by 1.
- Print the elements of the set with the help of iterator.
Explanation with Example:
Binary Tree -
50 --- Level 0
/ \
60 70 --- Level 1
/ \ / \
90 40 40 20 --- Level 2
K = 2
Initialize Queue and Set and append Root in queue
Step 1:
Queue = [50], Set = {}, Level = 0
As current Level is not equal to K,
Deque nodes from the queue and enqueue its child
Step 2:
Queue = [60, 70], Set = {}, Level = 1
As current level is not equal to K
Deque nodes one by one from the queue and enqueue its child
Step 3:
Queue = [90, 40, 40, 20], Set = {}, Level = 2
As the current level is equal to K
Deque all the nodes from the queue and add to the set
Set = {90, 40, 20}
Below is the implementation of the approach:
C++
// C++ implementation to print the // nodes of Kth Level without duplicates#include <bits/stdc++.h>using namespace std;// Structure of Binary Tree nodestruct node { int data; struct node* left; struct node* right;};// Function to create new// Binary Tree nodestruct node* newNode(int data){ struct node* temp = new struct node; temp->data = data; temp->left = nullptr; temp->right = nullptr; return temp;};// Function to print the nodes// of Kth Level without duplicatesvoid nodesAtKthLevel(struct node* root, int k){ // Condition to check if current // node is None if (root == nullptr) return; // Create Queue queue<struct node*> que; // Enqueue the root node que.push(root); // Create a set set<int> s; // Level to track // the current level int level = 0; int flag = 0; // Iterate the queue till its not empty while (!que.empty()) { // Calculate the number of nodes // in the current level int size = que.size(); // Process each node of the current // level and enqueue their left // and right child to the queue while (size--) { struct node* ptr = que.front(); que.pop(); // If the current level matches the // required level then add into set if (level == k) { // Flag initialized to 1 flag = 1; // Inserting node data in set s.insert(ptr->data); } else { // Traverse to the left child if (ptr->left) que.push(ptr->left); // Traverse to the right child if (ptr->right) que.push(ptr->right); } } // Increment the variable level // by 1 for each level level++; // Break out from the loop // if the Kth Level is reached if (flag == 1) break; } set<int>::iterator it; for (it = s.begin(); it != s.end(); ++it) { cout << *it << " "; } cout << endl;}// Driver codeint main(){ struct node* root = new struct node; // Tree Construction root = newNode(60); root->left = newNode(20); root->right = newNode(30); root->left->left = newNode(80); root->left->right = newNode(10); root->right->left = newNode(40); int level = 1; nodesAtKthLevel(root, level); return 0;} |
Java
// Java implementation to print the // nodes of Kth Level without duplicatesimport java.util.*;class GFG{ // Structure of Binary Tree nodestatic class node { int data; node left; node right;}; // Function to create new// Binary Tree nodestatic node newNode(int data){ node temp = new node(); temp.data = data; temp.left = null; temp.right = null; return temp;}; // Function to print the nodes// of Kth Level without duplicatesstatic void nodesAtKthLevel(node root, int k){ // Condition to check if current // node is None if (root == null) return; // Create Queue Queue<node> que = new LinkedList<node>(); // Enqueue the root node que.add(root); // Create a set HashSet<Integer> s = new HashSet<Integer>(); // Level to track // the current level int level = 0; int flag = 0; // Iterate the queue till its not empty while (!que.isEmpty()) { // Calculate the number of nodes // in the current level int size = que.size(); // Process each node of the current // level and enqueue their left // and right child to the queue while (size-- > 0) { node ptr = que.peek(); que.remove(); // If the current level matches the // required level then add into set if (level == k) { // Flag initialized to 1 flag = 1; // Inserting node data in set s.add(ptr.data); } else { // Traverse to the left child if (ptr.left!=null) que.add(ptr.left); // Traverse to the right child if (ptr.right!=null) que.add(ptr.right); } } // Increment the variable level // by 1 for each level level++; // Break out from the loop // if the Kth Level is reached if (flag == 1) break; } for (int it : s) { System.out.print(it+ " "); } System.out.println();} // Driver codepublic static void main(String[] args){ node root = new node(); // Tree Construction root = newNode(60); root.left = newNode(20); root.right = newNode(30); root.left.left = newNode(80); root.left.right = newNode(10); root.right.left = newNode(40); int level = 1; nodesAtKthLevel(root, level); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation to print the# nodes of Kth Level without duplicatesfrom collections import deque# A binary tree node has key, pointer to # left child and a pointer to right childclass Node: def __init__(self, key): self.data = key self.left = None self.right = None# Function to print the nodes# of Kth Level without duplicatesdef nodesAtKthLevel(root: Node, k: int): # Condition to check if current # node is None if root is None: return # Create Queue que = deque() # Enqueue the root node que.append(root) # Create a set s = set() # Level to track # the current level level = 0 flag = 0 # Iterate the queue till its not empty while que: # Calculate the number of nodes # in the current level size = len(que) # Process each node of the current # level and enqueue their left # and right child to the queue while size: ptr = que[0] que.popleft() # If the current level matches the # required level then add into set if level == k: # Flag initialized to 1 flag = 1 # Inserting node data in set s.add(ptr.data) else: # Traverse to the left child if ptr.left: que.append(ptr.left) # Traverse to the right child if ptr.right: que.append(ptr.right) size -= 1 # Increment the variable level # by 1 for each level level += 1 # Break out from the loop # if the Kth Level is reached if flag == 1: break for it in s: print(it, end = " ") print()# Driver Codeif __name__ == "__main__": # Tree Construction root = Node(60) root.left = Node(20) root.right = Node(30) root.left.left = Node(80) root.left.right = Node(10) root.right.left = Node(40) level = 1 nodesAtKthLevel(root, level)# This code is contributed by sanjeev2552 |
C#
// C# implementation to print the // nodes of Kth Level without duplicatesusing System;using System.Collections.Generic;class GFG{ // Structure of Binary Tree nodeclass node { public int data; public node left; public node right;}; // Function to create new// Binary Tree nodestatic node newNode(int data){ node temp = new node(); temp.data = data; temp.left = null; temp.right = null; return temp;} // Function to print the nodes// of Kth Level without duplicatesstatic void nodesAtKthLevel(node root, int k){ // Condition to check if current // node is None if (root == null) return; // Create Queue List<node> que = new List<node>(); // Enqueue the root node que.Add(root); // Create a set HashSet<int> s = new HashSet<int>(); // Level to track // the current level int level = 0; int flag = 0; // Iterate the queue till its not empty while (que.Count != 0) { // Calculate the number of nodes // in the current level int size = que.Count; // Process each node of the current // level and enqueue their left // and right child to the queue while (size-- > 0) { node ptr = que[0]; que.RemoveAt(0); // If the current level matches the // required level then add into set if (level == k) { // Flag initialized to 1 flag = 1; // Inserting node data in set s.Add(ptr.data); } else { // Traverse to the left child if (ptr.left != null) que.Add(ptr.left); // Traverse to the right child if (ptr.right != null) que.Add(ptr.right); } } // Increment the variable level // by 1 for each level level++; // Break out from the loop // if the Kth Level is reached if (flag == 1) break; } foreach (int it in s) { Console.Write(it+ " "); } Console.WriteLine();} // Driver codepublic static void Main(String[] args){ node root = new node(); // Tree Construction root = newNode(60); root.left = newNode(20); root.right = newNode(30); root.left.left = newNode(80); root.left.right = newNode(10); root.right.left = newNode(40); int level = 1; nodesAtKthLevel(root, level); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation to print the // nodes of Kth Level without duplicates// Structure of Binary Tree nodeclass node { constructor() { this.data = 0; this.left = null; this.right = null; }}; // Function to create new// Binary Tree nodefunction newNode(data){ var temp = new node(); temp.data = data; temp.left = null; temp.right = null; return temp;} // Function to print the nodes// of Kth Level without duplicatesfunction nodesAtKthLevel(root, k){ // Condition to check if current // node is None if (root == null) return; // Create Queue var que = []; // Enqueue the root node que.push(root); // Create a set var s = new Set(); // Level to track // the current level var level = 0; var flag = 0; // Iterate the queue till its not empty while (que.length != 0) { // Calculate the number of nodes // in the current level var size = que.length; // Process each node of the current // level and enqueue their left // and right child to the queue while (size-- > 0) { var ptr = que[0]; que.shift(); // If the current level matches the // required level then add into set if (level == k) { // Flag initialized to 1 flag = 1; // Inserting node data in set s.add(ptr.data); } else { // Traverse to the left child if (ptr.left != null) que.push(ptr.left); // Traverse to the right child if (ptr.right != null) que.push(ptr.right); } } // Increment the variable level // by 1 for each level level++; // Break out from the loop // if the Kth Level is reached if (flag == 1) break; } for(var it of s) { document.write(it+ " "); } document.write("<br>");} // Driver codevar root = new node();// Tree Constructionroot = newNode(60);root.left = newNode(20);root.right = newNode(30);root.left.left = newNode(80);root.left.right = newNode(10);root.right.left = newNode(40);var level = 1;nodesAtKthLevel(root, level);</script> |
Output:
20 30
Performance Analysis:
- Time Complexity: As in the above approach in the worst case all the N nodes of the Tree are visited, So the Time complexity will be O(N)
- Space Complexity: As in the worst case at the bottom most level of the Tree it can have the maximum number of the nodes which is 2H-1 where H is the height of the Binary Tree, then Space complexity of the Binary Tree will be O(2H-1)
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