# Nodes at Kth level without duplicates in a Binary Tree

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.
1. Find the size of the queue and store in a variable size to visit only the nodes of a current level.
2. Iterate with another while loop until the size variable is not 0
3. Deque a node from the queue and Enqueue its Left and right childs in the Queue.
4. 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 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 que;     // Enqueue the root node    que.push(root);     // Create a set    set 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::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 que = new LinkedList();      // Enqueue the root node    que.add(root);      // Create a set    HashSet s = new HashSet();      // 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 que = new List();       // Enqueue the root node    que.Add(root);       // Create a set    HashSet s = new HashSet();       // 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



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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