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 node struct node { int data; struct node* left; struct node* right; }; // Function to create new // Binary Tree node struct 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 duplicates void 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 code int 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 duplicates import java.util.*; class GFG{ // Structure of Binary Tree node static class node { int data; node left; node right; }; // Function to create new // Binary Tree node static 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 duplicates static 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 code public 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 duplicates from collections import deque # A binary tree node has key, pointer to # left child and a pointer to right child class Node: def __init__( self , key): self .data = key self .left = None self .right = None # Function to print the nodes # of Kth Level without duplicates def 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 Code if __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 duplicates using System; using System.Collections.Generic; class GFG{ // Structure of Binary Tree node class node { public int data; public node left; public node right; }; // Function to create new // Binary Tree node static 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 duplicates static 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 code public 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 node class node { constructor() { this .data = 0; this .left = null ; this .right = null ; } }; // Function to create new // Binary Tree node function 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 duplicates function 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 code var 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); 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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