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

Given a binary tree, the task is to print the odd 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: 3 9

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

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++

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// 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 printOddLevelOddNodes(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);
  
    printOddLevelOddNodes(root);
  
    return 0;
}

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Java














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// 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 printOddLevelOddNodes(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); 


  


    printOddLevelOddNodes(root); 


} 


}  


  


// This code is contributed by Princi Singh 



















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# 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 printOddLevelOddNodes(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 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) 


  


    printOddLevelOddNodes(root)  



















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C#














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// 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 printOddLevelOddNodes(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); 


  


    printOddLevelOddNodes(root); 


} 


} 


  


// This code is contributed by 29AjayKumar 



















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Output:


3 9






          
          
          
          

            

    

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