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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

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 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 

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 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)  

C#

// 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 

Output:

2 8

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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