Print all odd nodes of Binary Search Tree

Difficulty Level : Medium
Last Updated : 31 Oct, 2019

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 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

Output:

3 5 7

Time Complexity : O(n) where n is no. Of nodes

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