Print the nodes at odd levels of a tree

Difficulty Level : Easy
Last Updated : 17 May, 2021

Given a binary tree, print nodes of odd level in any order. Root is considered at level 1.

For example consider the following tree
          1
       /     \
      2       3
    /   \       \
   4     5       6
        /  \     /
       7    8   9

Output  1 4 5 6

Method 1 (Recursive)

The idea is to pass initial level as odd and switch level flag in every recursive call. For every node, if odd flag is set, then print it.

C++

// Recursive C++ program to print odd level nodes
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    Node* left, *right;
};
 
void printOddNodes(Node *root, bool isOdd = true)
{
    // If empty tree
    if (root == NULL)
      return;
 
    // If current node is of odd level
    if (isOdd)
       cout << root->data << " " ;
 
    // Recur for children with isOdd
    // switched.
    printOddNodes(root->left, !isOdd);
    printOddNodes(root->right, !isOdd);
}
 
// 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);
    printOddNodes(root);
 
    return 0;
}

Java

// Recursive Java program to print odd level nodes
class GfG {
 
static class Node {
    int data;
    Node left, right;
}
 
static void printOddNodes(Node root, boolean isOdd)
{
    // If empty tree
    if (root == null)
    return;
 
    // If current node is of odd level
    if (isOdd == true)
    System.out.print(root.data + " ");
 
    // Recur for children with isOdd
    // switched.
    printOddNodes(root.left, !isOdd);
    printOddNodes(root.right, !isOdd);
}
 
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    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);
    printOddNodes(root, true);
 
}
}

Python3

# Recursive Python3 program to print
# odd level nodes
 
# Utility method to create a node
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
 
def printOddNodes(root, isOdd = True):
     
    # If empty tree
    if (root == None):
        return
 
    # If current node is of odd level
    if (isOdd):
        print(root.data, end = " ")
 
    # Recur for children with isOdd
    # switched.
    printOddNodes(root.left, not isOdd)
    printOddNodes(root.right, not isOdd)
 
# 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)
    printOddNodes(root)
     
# This code is contributed by PranchalK

C#

using System;
 
// Recursive C# program to print odd level nodes 
public class GfG
{
 
public class Node
{
    public int data;
    public Node left, right;
}
 
public static void printOddNodes(Node root, bool isOdd)
{
    // If empty tree 
    if (root == null)
    {
    return;
    }
 
    // If current node is of odd level 
    if (isOdd == true)
    {
    Console.Write(root.data + " ");
    }
 
    // Recur for children with isOdd 
    // switched. 
    printOddNodes(root.left, !isOdd);
    printOddNodes(root.right, !isOdd);
}
 
// Utility method to create a node 
public static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    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);
    printOddNodes(root, true);
 
}
}
 
// This code is contributed by Shrikant13

Output:

1 4 5

Time complexity : O(n)

Method 2 (Iterative)

The above code prints nodes in preorder way. If we wish to print nodes level by level, we can use level order traversal. The idea is based on Print level order traversal line by line
We traverse nodes level by level. We switch odd level flag after every level.

C++

// Iterative C++ program to print odd level nodes
#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 printOddNodes(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 odd
    q.push(root);
    bool isOdd = true; 
 
    while (1)
    {
        // nodeCount (queue size) indicates
        // number of nodes at current level.
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Dequeue all nodes of current level
        // and Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            Node *node = q.front();
            if (isOdd)
               cout << node->data << " ";
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
        }
 
        isOdd = !isOdd;
    }
}
 
// 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);
    printOddNodes(root);
 
    return 0;
}

Java

// Iterative Java program to print odd level nodes
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 printOddNodes(Node root)
{
    // Base Case
    if (root == null) return;
 
    // Create an empty queue for level
    // order traversal
    Queue<Node> q = new LinkedList<Node> ();
 
    // Enqueue root and initialize level as odd
    q.add(root);
    boolean isOdd = true;
 
    while (true)
    {
        // nodeCount (queue size) indicates
        // number of nodes at current level.
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Dequeue all nodes of current level
        // and Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            Node node = q.peek();
            if (isOdd == true)
            System.out.print(node.data + " ");
            q.remove();
            if (node.left != null)
                q.add(node.left);
            if (node.right != null)
                q.add(node.right);
            nodeCount--;
        }
 
        isOdd = !isOdd;
    }
}
 
// Utility method to create a node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    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);
    printOddNodes(root);
}
}

Python3

# Iterative Python3 program to prodd
# level nodes
 
# A Binary Tree Node
# Utility function to create a
# new tree Node
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None
         
# Iterative method to do level order
# traversal line by line
def printOddNodes(root) :
 
    # Base Case
    if (root == None):
        return
 
    # Create an empty queue for 
    # level order traversal
    q = []
 
    # Enqueue root and initialize
    # level as odd
    q.append(root)
    isOdd = True
 
    while (1) :
     
        # nodeCount (queue size) indicates
        # number of nodes at current level.
        nodeCount = len(q)
        if (nodeCount == 0) :
            break
 
        # Dequeue all nodes of current level
        # and Enqueue all nodes of next level
        while (nodeCount > 0):
         
            node = q[0]
            if (isOdd):
                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
             
        isOdd = not isOdd
 
# 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)
    printOddNodes(root)
 
# This code is contributed
# by SHUBHAMSINGH10

C#

// Iterative C# program to
// print odd level nodes
using System;
using System.Collections.Generic;
 
public class GfG
{
    public class Node
    {
        public int data;
        public Node left, right;
    }
 
    // Iterative method to do level
    // order traversal line by line
    static void printOddNodes(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 odd
        q.Enqueue(root);
        bool isOdd = true;
 
        while (true)
        {
            // nodeCount (queue size) indicates
            // number of nodes at current level.
            int nodeCount = q.Count;
            if (nodeCount == 0)
                break;
 
            // Dequeue all nodes of current level
            // and Enqueue all nodes of next level
            while (nodeCount > 0)
            {
                Node node = q.Peek();
                if (isOdd == true)
                    Console.Write(node.data + " ");
                q.Dequeue();
                if (node.left != null)
                    q.Enqueue(node.left);
                if (node.right != null)
                    q.Enqueue(node.right);
                nodeCount--;
            }
            isOdd = !isOdd;
        }
    }
 
    // Utility method to create a node
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = null;
        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);
        printOddNodes(root);
    }
}
 
// This code has been contributed
// by 29AjayKumar

Output:

1 4 5

Time complexity : O(n)

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