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

Given a binary tree, the task is to print the even positioned nodes of odd 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: 2 8

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

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 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 printOddLevelEvenNodes(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->left->left = newNode(10);
    root->left->right->right->right = newNode(11);
  
    printOddLevelEvenNodes(root);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 printOddLevelEvenNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
  
    // Create an empty queue for level
    // order traversal
    Queue<Node> 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.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
  
    printOddLevelEvenNodes(root);
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 printOddLevelEvenNodes(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 not 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.left.left = newNode(10)
    root.left.right.right.right = newNode(11)
  
    printOddLevelEvenNodes(root) 

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
class Node
{
    public int data;
    public Node left, right;
};
  
// Iterative method to do level order
// traversal line by line
static void printOddLevelEvenNodes(Node root)
{
    // Base Case
    if (root == null)
        return;
  
    // Create an empty queue for level
    // order traversal
    Queue<Node> q = new Queue<Node>();
  
    // 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.left.left = newNode(10);
    root.left.right.right.right = newNode(11);
  
    printOddLevelEvenNodes(root);
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

2 8


My Personal Notes arrow_drop_up

Small things always make you to think big

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : 29AjayKumar



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.