Print even positioned nodes of even levels in level order of the given binary tree

Given a binary tree, print even positioned nodes of even level in level order traversal. The root is considered at level 0, and the left most node of any level is considered as a node at position 0.

Examples:

Input:
1
/   \
2     3
/   \     \
4     5      6
/  \
7    8
/      \
9        10

Output: 1 4 6 9

Input:
2
/   \
4     15
/     /
45   17

Output: 2 45

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To print nodes level by level, use level order traversal. The idea is based on Print level order traversal line by line. For that, traverse nodes level by level and switch even level flag after every level. Similarly, mark 1st node in every level as even position and switch it after each time the next node is processed.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    struct Node {     int data;     Node *left, *right; };    // Iterative method to do level order // traversal line by line void printEvenLevelEvenNodes(Node* root) {     // Base Case     if (root == NULL)         return;        // Create an empty queue for level     // order traversal     queue q;        // Enqueue root and initialize level as even     q.push(root);     bool evenLevel = true;        while (1) {            // nodeCount (queue size) indicates         // number of nodes in the current level         int nodeCount = q.size();         if (nodeCount == 0)             break;            // Mark 1st node as even positioned         bool evenNodePosition = true;            // Dequeue all the nodes of current level         // and Enqueue all the nodes of next level         while (nodeCount > 0) {             Node* node = q.front();                // Print only even positioned             // nodes of even levels             if (evenLevel && evenNodePosition)                 cout << node->data << " ";             q.pop();             if (node->left != NULL)                 q.push(node->left);             if (node->right != NULL)                 q.push(node->right);             nodeCount--;                // Switch the even position flag             evenNodePosition = !evenNodePosition;         }            // Switch the even level flag         evenLevel = !evenLevel;     } }    // Utility method to create a node struct Node* newNode(int data) {     struct Node* node = new Node;     node->data = data;     node->left = node->right = NULL;     return (node); }    // Driver code int main() {     struct Node* root = newNode(1);     root->left = newNode(2);     root->right = newNode(3);     root->left->left = newNode(4);     root->left->right = newNode(5);     root->right->left = newNode(6);     root->right->right = newNode(7);     root->left->right->left = newNode(8);     root->left->right->right = newNode(9);     root->left->right->right->right = newNode(10);        printEvenLevelEvenNodes(root);        return 0; }

Java

 // Java implementation of the approach import java.util.*;    class GFG  {    static class Node  {     int data;     Node left, right; };    // Iterative method to do level order // traversal line by line static void printEvenLevelEvenNodes(Node root) {     // Base Case     if (root == null)         return;        // Create an empty queue for level     // order traversal     Queue q = new LinkedList();        // Enqueue root and initialize level as even     q.add(root);     boolean evenLevel = true;        while (true)     {            // nodeCount (queue size) indicates         // number of nodes in the current level         int nodeCount = q.size();         if (nodeCount == 0)             break;            // Mark 1st node as even positioned         boolean evenNodePosition = true;            // Dequeue all the nodes of current level         // and Enqueue all the nodes of next level         while (nodeCount > 0)         {             Node node = q.peek();                // Print only even positioned             // nodes of even levels             if (evenLevel && evenNodePosition)                 System.out.print(node.data + " ");             q.remove();             if (node.left != null)                 q.add(node.left);             if (node.right != null)                 q.add(node.right);             nodeCount--;                // Switch the even position flag             evenNodePosition = !evenNodePosition;         }            // Switch the even level flag         evenLevel = !evenLevel;     } }    // Utility method to create a node static Node newNode(int data) {     Node node = new Node();     node.data = data;     node.left = node.right = null;     return (node); }    // Driver code public static void main(String[] args)  {     Node root = newNode(1);     root.left = newNode(2);     root.right = newNode(3);     root.left.left = newNode(4);     root.left.right = newNode(5);     root.right.left = newNode(6);     root.right.right = newNode(7);     root.left.right.left = newNode(8);     root.left.right.right = newNode(9);     root.left.right.right.right = newNode(10);        printEvenLevelEvenNodes(root); } }     // This code is contributed by 29AjayKumar

Python3

 # Python3 implementation of the approach     # Utility method to create a node  class newNode:         # Construct to create a new node      def __init__(self, key):          self.data = key          self.left = None         self.right = None    # Iterative method to do level order  # traversal line by line  def printEvenLevelEvenNodes(root):            # Base Case      if (root == None):         return            # Create an empty queue for level      # order traversal      q = []             # Enqueue root and initialize     # level as even     q.append(root)      evenLevel = True            while (1):                    # nodeCount (queue size) indicates          # number of nodes in the current level          nodeCount = len(q)         if (nodeCount == 0):             break                    # Mark 1st node as even positioned          evenNodePosition = True                    # Dequeue all the nodes of current level          # and Enqueue all the nodes of next level          while (nodeCount > 0):             node = q                            # Pronly even positioned              # nodes of even levels              if (evenLevel and evenNodePosition):                 print(node.data, end = " ")             q.pop(0)             if (node.left != None):                 q.append(node.left)             if (node.right != None):                 q.append(node.right)              nodeCount -= 1                            # Switch the even position flag              evenNodePosition = not evenNodePosition                     # Switch the even level flag          evenLevel = not evenLevel         # Driver code  if __name__ == '__main__':             root = newNode(1)      root.left = newNode(2)      root.right = newNode(3)      root.left.left = newNode(4)      root.left.right = newNode(5)      root.right.left = newNode(6)      root.right.right = newNode(7)      root.left.right.left = newNode(8)      root.left.right.right = newNode(9)      root.left.right.right.right = newNode(10)         printEvenLevelEvenNodes(root)     # This code is contributed by SHUBHAMSINGH10

C#

 // C# implementation of the approach using System; using System.Collections.Generic;    class GFG  { public class Node  {     public int data;     public Node left, right; };    // Iterative method to do level order // traversal line by line static void printEvenLevelEvenNodes(Node root) {     // Base Case     if (root == null)         return;        // Create an empty queue for level     // order traversal     Queue q = new Queue ();        // Enqueue root and initialize level as even     q.Enqueue(root);     bool evenLevel = true;        while (true)     {            // nodeCount (queue size) indicates         // number of nodes in the current level         int nodeCount = q.Count;         if (nodeCount == 0)             break;            // Mark 1st node as even positioned         bool evenNodePosition = true;            // Dequeue all the nodes of current level         // and Enqueue all the nodes of next level         while (nodeCount > 0)         {             Node node = q.Peek();                // Print only even positioned             // nodes of even levels             if (evenLevel && evenNodePosition)                 Console.Write(node.data + " ");             q.Dequeue();             if (node.left != null)                 q.Enqueue(node.left);             if (node.right != null)                 q.Enqueue(node.right);             nodeCount--;                // Switch the even position flag             evenNodePosition = !evenNodePosition;         }            // Switch the even level flag         evenLevel = !evenLevel;     } }    // Utility method to create a node static Node newNode(int data) {     Node node = new Node();     node.data = data;     node.left = node.right = null;     return (node); }    // Driver code public static void Main(String[] args)  {     Node root = newNode(1);     root.left = newNode(2);     root.right = newNode(3);     root.left.left = newNode(4);     root.left.right = newNode(5);     root.right.left = newNode(6);     root.right.right = newNode(7);     root.left.right.left = newNode(8);     root.left.right.right = newNode(9);     root.left.right.right.right = newNode(10);        printEvenLevelEvenNodes(root); } }     // This code is contributed by PrinciRaj1992

Output:

1 4 6 10

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