Print the nodes at odd levels of a tree

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.

Implementation:

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

Javascript

<script>

    // Recursive JavaScript program to print odd level nodes

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    function printOddNodes(root, isOdd)
    {
        // If empty tree
        if (root == null)
        return;

        // If current node is of odd level
        if (isOdd == true)
        document.write(root.data + " ");

        // Recur for children with isOdd
        // switched.
        printOddNodes(root.left, !isOdd);
        printOddNodes(root.right, !isOdd);
    }

    // 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);
    printOddNodes(root, true);

</script>

Output

1 4 5 

Time complexity : O(n)
Auxiliary Space: O(n) for Recursive Stack Space in case of Skewed Tree

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. 

Implementation:

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 print odd
# 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

Javascript

<script>
// Iterative Javascript program to print odd level nodes

class Node
{
    constructor(data)
    {
        this.data=data;
        this.left=this.right=null;
    }
}

function printOddNodes(root)
{
    // Base Case
    if (root == null) return;

    // Create an empty queue for level
    // order traversal
    let q = [];

    // Enqueue root and initialize level as odd
    q.push(root);
    let isOdd = true;

    while (true)
    {
        // nodeCount (queue size) indicates
        // number of nodes at current level.
        let nodeCount = q.length;
        if (nodeCount == 0)
            break;

        // Dequeue all nodes of current level
        // and Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            let node = q[0];
            if (isOdd == true)
                document.write(node.data + " ");
            q.shift();
            if (node.left != null)
                q.push(node.left);
            if (node.right != null)
                q.push(node.right);
            nodeCount--;
        }

        isOdd = !isOdd;
    }
}

// Driver code
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
printOddNodes(root);

// This code is contributed by rag2127
</script>

Output

1 4 5 

Time complexity : O(n) 
Auxiliary Space: O(n) for Queue Data Structure

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