Print all odd nodes of Binary Search Tree

07 Feb, 2022
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++
Java
Python3
C#
Javascript
// C++ program to print all odd node of BST#include <bits/stdc++.h>using namespace std; // create Treestruct Node {    int key;    struct Node *left, *right;}; // A utility function to create a new BST nodeNode* 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 BSTvoid 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 nodesvoid 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 Codeint 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 program to print all odd node of BSTclass GfG { // create Treestatic class Node {    int key;    Node left, right;} // A utility function to create a new BST nodestatic 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 BSTstatic 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 nodewith 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 nodesstatic 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 Codepublic 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 program to print all odd# node of BST # create Tree# to create a new BST nodeclass 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 BSTdef inorder( root) :     if (root != None):        inorder(root.left)        print(root.key, end = " ")        inorder(root.right)     """ A utility function to insert anew 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 nodesdef 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 Codeif __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# program to print all odd node of BSTusing System; public class GfG{ // create Treeclass Node{    public int key;    public Node left, right;} // A utility function to create a new BST nodestatic 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 BSTstatic void inorder(Node root){    if (root != null)    {        inorder(root.left);        Console.Write(root.key + " ");        inorder(root.right);    }} /* A utility function to insert a new nodewith 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 nodesstatic 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 Codepublic 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
<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>
3 5 7
 
Time Complexity : O(n) where n is no. of nodes in the binary search tree.Auxiliary Space: O(n) 
prerna saini
VishalBachchas
princiraj1992
SHUBHAMSINGH10
nidhi_biet
pankajsharmagfg
Rajput-Ji
sweetyty
Technical Scripter 2018
Binary Search Tree
Data Structures
Technical Scripter
Data Structures
Binary Search Tree
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