# 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 10Output:1 4 6 9Input:2 / \ 4 15 / / 45 17Output: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++ implementation of the approach ` `#include <bits/stdc++.h> ` `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<Node*> 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1 4 6 10

## Recommended Posts:

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- Recursive Program to Print extreme nodes of each level of Binary Tree in alternate order
- Print all nodes between two given levels in Binary Tree
- Print Levels of all nodes in a Binary Tree
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- Print Binary Tree levels in sorted order | Set 2 (Using set)
- Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)
- Print nodes between two given level numbers of a binary tree
- Print all the nodes except the leftmost node in every level of the given binary tree
- Print the nodes at odd levels of a tree
- Difference between sums of odd level and even level nodes of a Binary Tree
- Flatten Binary Tree in order of Level Order Traversal
- Maximum sum of leaf nodes among all levels of the given binary tree
- Maximum sum of non-leaf nodes among all levels of the given binary tree

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