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# 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```

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[``0``]``            ` `            ``# Print only 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`

## Javascript

 ``

Output:

`1 4 6 10`

Time Complexity: O(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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