Print all odd nodes of Binary Search Tree

Given a binary search tree. The task is to print all odd nodes of the binary search tree.

Examples: 

Input : 
          5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8 
Output : 3 5 7

Input :
          14 
        /   \ 
       12    17 
      / \   / \ 
     8  13 16   19 
Output : 13 17 19

Approach: Traverse the Binary Search tree using any of the tree traversals and check if the current node’s value is odd. If yes then print it otherwise skip that node.

Below is the implementation of the above Approach: 

C++

// C++ program to print all odd node of BST
#include <bits/stdc++.h>
using namespace std;

// create Tree
struct Node {
    int key;
    struct Node *left, *right;
};

// A utility function to create a new BST node
Node* newNode(int item)
{
    Node* temp = new Node;
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}

// A utility function to do inorder traversal of BST
void inorder(Node* root)
{
    if (root != NULL) {
        inorder(root->left);
        printf("%d ", root->key);
        inorder(root->right);
    }
}

/* A utility function to insert a new node
   with given key in BST */
Node* insert(Node* node, int key)
{
    /* If the tree is empty, return a new node */
    if (node == NULL)
        return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node->key)
        node->left = insert(node->left, key);
    else
        node->right = insert(node->right, key);

    /* return the (unchanged) node pointer */
    return node;
}

// Function to print all odd nodes
void oddNode(Node* root)
{
    if (root != NULL) {
        oddNode(root->left);

        // if node is odd then print it
        if (root->key % 2 != 0)
            printf("%d ", root->key);

        oddNode(root->right);
    }
}

// Driver Code
int main()
{
    /* Let us create following BST
        5
      /  \
     3    7
    / \  / \
    2 4  6 8 */
    Node* root = NULL;
    root = insert(root, 5);
    root = insert(root, 3);
    root = insert(root, 2);
    root = insert(root, 4);
    root = insert(root, 7);
    root = insert(root, 6);
    root = insert(root, 8);

    oddNode(root);

    return 0;
}

Java

// Java program to print all odd node of BST
class GfG {

// create Tree
static class Node {
    int key;
    Node left, right;
}

// A utility function to create a new BST node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.key = item;
    temp.left = null;
    temp.right = null;
    return temp;
}

// A utility function to do inorder traversal of BST
static void inorder(Node root)
{
    if (root != null) {
        inorder(root.left);
        System.out.print(root.key + " ");
        inorder(root.right);
    }
}

/* A utility function to insert a new node
with given key in BST */
static Node insert(Node node, int key)
{
    /* If the tree is empty, return a new node */
    if (node == null)
        return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node.key)
        node.left = insert(node.left, key);
    else
        node.right = insert(node.right, key);

    /* return the (unchanged) node pointer */
    return node;
}

// Function to print all odd nodes
static void oddNode(Node root)
{
    if (root != null) {
        oddNode(root.left);

        // if node is odd then print it
        if (root.key % 2 != 0)
            System.out.print(root.key + " ");

        oddNode(root.right);
    }
}

// Driver Code
public static void main(String[] args)
{
    /* Let us create following BST
        5
    / \
    3 7
    / \ / \
    2 4 6 8 */
    Node root = null;
    root = insert(root, 5);
    root = insert(root, 3);
    root = insert(root, 2);
    root = insert(root, 4);
    root = insert(root, 7);
    root = insert(root, 6);
    root = insert(root, 8);

    oddNode(root);

}
}

Python3

# Python3 program to print all odd
# node of BST

# create Tree
# to create a new BST node
class newNode:

    # Construct to create a new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None

# A utility function to do inorder
# traversal of BST
def inorder( root) :

    if (root != None):
        inorder(root.left)
        print(root.key, end = " ")
        inorder(root.right)

""" A utility function to insert a
new node with given key in BST """
def insert(node, key):

    """ If the tree is empty,
    return a new node """
    if (node == None):
        return newNode(key)

    """ Otherwise, recur down the tree """
    if (key < node.key):
        node.left = insert(node.left, key)
    else:
        node.right = insert(node.right, key)

    """ return the (unchanged) node pointer """
    return node

# Function to print all even nodes
def oddNode(root) :

    if (root != None):
        oddNode(root.left)

        # if node is even then print it
        if (root.key % 2 != 0):
            print(root.key, end = " ")
        oddNode(root.right)

# Driver Code
if __name__ == '__main__':

    """ Let us create following BST
    5
    / \
    3 7
    / \ / \
    2 4 6 8 """
    root = None
    root = insert(root, 5)
    root = insert(root, 3)
    root = insert(root, 2)
    root = insert(root, 4)
    root = insert(root, 7)
    root = insert(root, 6)
    root = insert(root, 8)

    oddNode(root)

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to print all odd node of BST
using System;

public class GfG
{

// create Tree
class Node
{
    public int key;
    public Node left, right;
}

// A utility function to create a new BST node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.key = item;
    temp.left = null;
    temp.right = null;
    return temp;
}

// A utility function to do
// inorder traversal of BST
static void inorder(Node root)
{
    if (root != null)
    {
        inorder(root.left);
        Console.Write(root.key + " ");
        inorder(root.right);
    }
}

/* A utility function to insert a new node
with given key in BST */
static Node insert(Node node, int key)
{
    /* If the tree is empty, return a new node */
    if (node == null)
        return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node.key)
        node.left = insert(node.left, key);
    else
        node.right = insert(node.right, key);

    /* return the (unchanged) node pointer */
    return node;
}

// Function to print all odd nodes
static void oddNode(Node root)
{
    if (root != null)
    {
        oddNode(root.left);

        // if node is odd then print it
        if (root.key % 2 != 0)
            Console.Write(root.key + " ");

        oddNode(root.right);
    }
}

// Driver Code
public static void Main(String[] args)
{
    /* Let us create following BST
        5
    / \
    3 7
    / \ / \
    2 4 6 8 */
    Node root = null;
    root = insert(root, 5);
    root = insert(root, 3);
    root = insert(root, 2);
    root = insert(root, 4);
    root = insert(root, 7);
    root = insert(root, 6);
    root = insert(root, 8);

    oddNode(root);

}
}

// This code has been contributed
// by PrinciRaj1992

Javascript

<script>
// javascript program to print all odd node of BST

    // create Tree
     class Node {
         constructor(){
        this.key = 0;
        this.left = this.right = null;
    }
     }

    // A utility function to create a new BST node
    function newNode(item) {
        var temp = new Node();
        temp.key = item;
        temp.left = null;
        temp.right = null;
        return temp;
    }

    // A utility function to do inorder traversal of BST
    function inorder(root) {
        if (root != null) {
            inorder(root.left);
            document.write(root.key + " ");
            inorder(root.right);
        }
    }

    /*
     * A utility function to insert a new node with given key in BST
     */
    function insert(node, key)
    {

        /* If the tree is empty, return a new node */
        if (node == null)
            return newNode(key);

        /* Otherwise, recur down the tree */
        if (key < node.key)
            node.left = insert(node.left, key);
        else
            node.right = insert(node.right, key);

        /* return the (unchanged) node pointer */
        return node;
    }

    // Function to print all odd nodes
    function oddNode(root) {
        if (root != null) {
            oddNode(root.left);

            // if node is odd then print it
            if (root.key % 2 != 0)
                document.write(root.key + " ");

            oddNode(root.right);
        }
    }

    // Driver Code

        /*
         * Let us create following BST
     5
    / \
    3 7
    / \ / \
    2 4 6 8 */
        var root = null;
        root = insert(root, 5);
        root = insert(root, 3);
        root = insert(root, 2);
        root = insert(root, 4);
        root = insert(root, 7);
        root = insert(root, 6);
        root = insert(root, 8);

        oddNode(root);

// This code is contributed by Rajput-Ji
</script>

Output

3 5 7 

Complexity Analysis:

• Time Complexity: O(n) where n is no. of nodes in the binary search tree.
• Auxiliary Space: O(n)