Open In App

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

Given a binary tree, the task is to print the odd positioned nodes of even levels in the level order traversal of the tree. The root is considered at level 0, and the leftmost node of any level is considered as a node at position 0.
Example:

Input:
1
/    \
2       3
/ \      /  \
4   5    6    7
/  \
8    9
/      \
10       11
Output: 5 7 11

Input:
2
/   \
4     15
/     /
45   17
Output: 17

Prerequisite – Even positioned elements at even level
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 odd level flag after every level. Similarly, mark 2nd node in every level as odd 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 linevoid printEvenLevelOddNodes(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 nodestruct Node* newNode(int data){    struct Node* node = new Node;    node->data = data;    node->left = node->right = NULL;    return (node);} // Driver codeint 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->left->left = newNode(10);    root->left->right->right->right = newNode(11);     printEvenLevelOddNodes(root);     return 0;}

## Java

 // Java implementation of the above approachimport java.util.*;class GFG{ static class Node{    int data;    Node left, right;}; // Iterative method to do level order// traversal line by linestatic void printEvenLevelOddNodes(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 nodestatic Node newNode(int data){    Node node = new Node();    node.data = data;    node.left = node.right = null;    return (node);} // Driver codepublic 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.left.left = newNode(10);    root.left.right.right.right = newNode(11);     printEvenLevelOddNodes(root);}} // This code is contributed by Princi Singh

## Python3

 # Python implementation of the approach # Utility method to create a nodeclass 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 linedef printEvenLevelOddNodes(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]            # Pronly even positioned            # nodes of even levels            if evenLevel and not 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 codeif __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.left.left = newNode(10)    root.left.right.right.right = newNode(11)     printEvenLevelOddNodes(root)

## C#

 // C# implementation of the above approachusing 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 linestatic void printEvenLevelOddNodes(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 nodestatic Node newNode(int data){    Node node = new Node();    node.data = data;    node.left = node.right = null;    return (node);} // Driver codepublic 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.left.left = newNode(10);    root.left.right.right.right = newNode(11);     printEvenLevelOddNodes(root);}} // This code is contributed by 29AjayKumar

## Javascript



Output:

5 7 11

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.