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

Last Updated : 23 Jan, 2023

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 <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    Node *left, *right;
};
 
// Iterative method to do level order
// traversal line by line
void printEvenLevelOddNodes(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);
 
    printEvenLevelOddNodes(root);
 
    return 0;
}
 
  

Java

   // Java implementation of the above 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 printEvenLevelOddNodes(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);
 
    printEvenLevelOddNodes(root);
}
} 
 
// This code is contributed by Princi Singh
 
  

Python3

   # Python 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 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 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)
 
    printEvenLevelOddNodes(root) 
 
  

C#

   // C# implementation of the above 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 printEvenLevelOddNodes(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);
 
    printEvenLevelOddNodes(root);
}
}
 
// This code is contributed by 29AjayKumar
 
  

Javascript

   <script>
 
    // JavaScript implementation of the above approach
     
    class Node 
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Iterative method to do level order
    // traversal line by line
    function printEvenLevelOddNodes(root)
    {
        // Base Case
        if (root == null)
            return;
 
        // Create an empty queue for level
        // order traversal
        let q = [];
 
        // Enqueue root and initialize level as even
        q.push(root);
        let evenLevel = true;
 
        while (true)
        {
 
            // nodeCount (queue size) indicates
            // number of nodes in the current level
            let nodeCount = q.length;
            if (nodeCount == 0)
                break;
 
            // Mark 1st node as even positioned
            let evenNodePosition = true;
 
            // Dequeue all the nodes of current level
            // and Enqueue all the nodes of next level
            while (nodeCount > 0) 
            {
                let node = q[0];
 
                // Print only even positioned
                // nodes of even levels
                if (evenLevel && !evenNodePosition)
                    document.write(node.data + " ");
 
                q.shift();
                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
    function newNode(data)
    {
        let node = new Node(data);
        return (node);
    }
     
    let 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);
 
</script>
 
  

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.

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

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

C++

Java

Python3

C#

Javascript

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