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Special two digit numbers in a Binary Search Tree
  • Difficulty Level : Medium
  • Last Updated : 31 Dec, 2020

Given a Binary Search Trees, the task is to count the number of nodes which are having special two-digit numbers.
Prerequisite : Special Two Digit Number | Binary Search Tree 

Examples :  

Input : 15 7 987 21
Output : 0

Input : 19 99 57 1 22
Output : 2

Algorithm: Iterate through each node of tree recursively with a variable count, and check each node’s data for a special two-digit number. If it is then increment the variable count. In the end, return count.

C




// C program to count number of nodes in
// BST containing two digit special number
#include <stdio.h>
#include <stdlib.h>
 
// A Tree node
struct Node
{
    struct Node *left;
    int info;
    struct Node *right;
};
 
// Function to create a new node
void insert(struct Node **rt, int key)
{
    if(*rt == NULL)
    {
        (*rt) = (struct Node *)malloc(sizeof(struct Node));
        (*rt) -> left = NULL;
        (*rt) -> right = NULL;
        (*rt) -> info = key;
    }
    else if(key < ((*rt) -> info))
        insert(&((*rt) -> left), key);
    else
        insert(&(*rt) -> right, key);
}
 
// Function to find if number
// is special or not
int check(int num)
{
    int sum = 0, i = num, sum_of_digits, prod_of_digits ;
     
    // Check if number is two digit or not
    if(num < 10 || num > 99)
        return 0;
    else
    {
        sum_of_digits = (i % 10) + (i / 10);
        prod_of_digits = (i % 10) * (i / 10);
        sum = sum_of_digits + prod_of_digits;
    }
     
    if(sum == num)
        return 1;
    else
        return 0;
}
 
// Function to count number of special two digit number
void countSpecialDigit(struct Node *rt, int *c)
{
    int x;
    if(rt == NULL)
        return;
    else
    {
        x = check(rt -> info);
        if(x == 1)
            *c = *c + 1;
        countSpecialDigit(rt -> left, c);
        countSpecialDigit(rt -> right, c);
    }
}
 
// Driver program to test
int main()
{
    struct Node *root = NULL;
     
    // Initialize result
    int count = 0;
     
    // Function call to insert() to insert nodes
    insert(&root, 50);
    insert(&root, 29);
    insert(&root, 59);
    insert(&root, 19);
    insert(&root, 53);
    insert(&root, 556);
    insert(&root, 56);
    insert(&root, 94);
    insert(&root, 13);
 
    // Function call, to check each node for
    // special two digit number
    countSpecialDigit(root, &count);
    printf("%d", count);
     
    return 0;
}

Java




// Java program to count number of nodes in
// BST containing two digit special number
 
// A binary tree node
class Node
{
    int info;
    Node left, right;
 
    Node(int d)
    {
        info = d;
        left = right = null;
    }
}
 
class BinaryTree{
 
static Node head;
static int count;
 
// Function to create a new node
Node insert(Node node, int info)
{
     
    // If the tree is empty, return a new,
    // single node
    if (node == null)
    {
        return (new Node(info));
    }
    else
    {
         
        // Otherwise, recur down the tree
        if (info <= node.info)
        {
            node.left = insert(node.left, info);
        }
        else
        {
            node.right = insert(node.right, info);
        }
 
        // return the (unchanged) node pointer
        return node;
    }
}
 
// Function to find if number
// is special or not
static int check(int num)
{
    int sum = 0, i = num,
        sum_of_digits,
        prod_of_digits;
 
    // Check if number is two digit or not
    if (num < 10 || num > 99)
        return 0;
    else
    {
        sum_of_digits = (i % 10) + (i / 10);
        prod_of_digits = (i % 10) * (i / 10);
        sum = sum_of_digits + prod_of_digits;
    }
 
    if (sum == num)
        return 1;
    else
        return 0;
}
 
// Function to count number of special
// two digit number
static void countSpecialDigit(Node rt)
{
    int x;
    if (rt == null)
        return;
    else
    {
        x = check(rt.info);
        if (x == 1)
            count = count + 1;
             
        countSpecialDigit(rt.left);
        countSpecialDigit(rt.right);
    }
}
 
// Driver code
public static void main(String[] args)
{
    BinaryTree tree = new BinaryTree();
    Node root = null;
     
    root = tree.insert(root, 50);
    tree.insert(root, 29);
    tree.insert(root, 59);
    tree.insert(root, 19);
    tree.insert(root, 53);
    tree.insert(root, 556);
    tree.insert(root, 56);
    tree.insert(root, 94);
    tree.insert(root, 13);
     
    // Function call
    countSpecialDigit(root);
    System.out.println(count);
}
}
 
// This code is contributed by tushar_bansal

Python3




# Python3 program to count number of nodes in
# BST containing two digit special number
 
# A Tree node
class Node:
     
    def __init__(self, x):
         
        self.data = x
        self.left = None
        self.right = None
 
# Function to create a new node
def insert(node, data):
     
    global succ
     
    # If the tree is empty, return
    # a new node
    root = node
 
    if (node == None):
        return Node(data)
 
    # If key is smaller than root's key,
    # go to left subtree and set successor
    # as current node
    if (data < node.data):
        root.left = insert(node.left, data)
 
    # Go to right subtree
    elif (data > node.data):
        root.right = insert(node.right, data)
         
    return root
 
# Function to find if number
# is special or not
def check(num):
     
    sum = 0
    i = num
    #sum_of_digits, prod_of_digits
 
    # Check if number is two digit or not
    if (num < 10 or num > 99):
        return 0
    else:
        sum_of_digits = (i % 10) + (i // 10)
        prod_of_digits = (i % 10) * (i // 10)
        sum = sum_of_digits + prod_of_digits
 
    if (sum == num):
        return 1
    else:
        return 0
 
# Function to count number of special
# two digit number
def countSpecialDigit(rt):
     
    global c
     
    if (rt == None):
        return
    else:
        x = check(rt.data)
        if (x == 1):
            c += 1
             
        countSpecialDigit(rt.left)
        countSpecialDigit(rt.right)
 
# Driver code
if __name__ == '__main__':
     
    root = None
 
    # Initialize result
    c = 0
 
    # Function call to insert() to
    # insert nodes
    root = insert(root, 50)
    root = insert(root, 29)
    root = insert(root, 59)
    root = insert(root, 19)
    root = insert(root, 53)
    root = insert(root, 556)
    root = insert(root, 56)
    root = insert(root, 94)
    root = insert(root, 13)
 
    # Function call, to check each node
    # for special two digit number
    countSpecialDigit(root)
     
    print(c)
 
# This code is contributed by mohit kumar 29

C#




// C# program to count number of nodes in
// BST containing two digit special number
 
// A binary tree node
using System;
public class Node
{
  public int info;
  public Node left, right; 
  public Node(int d)
  {
    info = d;
    left = right = null;
  }
}
public class BinaryTree
{
  public static Node head;
  public static int count;
 
  // Function to create a new node
  public Node insert(Node node, int info)
  {
 
    // If the tree is empty, return a new,
    // single node
    if(node == null)
    {
      return (new Node(info));
    }
    else
    {
 
      // Otherwise, recur down the tree
      if(info <= node.info)
      {
        node.left = insert(node.left, info);
      }
      else
      {
        node.right = insert(node.right, info);
      }
    }
 
    // return the (unchanged) node pointer
    return node;
  }
 
  // Function to find if number
  // is special or not
  static int check(int num)
  {
    int sum = 0, i = num, sum_of_digits, prod_of_digits;
 
    // Check if number is two digit or not
    if(num < 10 || num > 99)
    {
      return 0;
    }
    else
    {
      sum_of_digits = (i % 10) + (i / 10);
      prod_of_digits = (i % 10) * (i / 10);
      sum = sum_of_digits + prod_of_digits;
    }
    if(sum == num)
    {
      return 1;
    }
    else
    {
      return 0;
    }
  }
 
  // Function to count number of special
  // two digit number
  static void countSpecialDigit(Node rt)
  {
    int x;
    if(rt == null)
    {
      return;
    }
    else
    {
      x = check(rt.info);
      if(x == 1)
      {
        count = count + 1;
 
      }
      countSpecialDigit(rt.left);
      countSpecialDigit(rt.right);
    }
  }
 
  // Driver code
  static public void Main ()
  {
    BinaryTree tree = new BinaryTree();
    Node root = null;
    root = tree.insert(root, 50);
    tree.insert(root, 29);
    tree.insert(root, 59);
    tree.insert(root, 19);
    tree.insert(root, 53);
    tree.insert(root, 556);
    tree.insert(root, 56);
    tree.insert(root, 94);
    tree.insert(root, 13);
 
    // Function call
    countSpecialDigit(root);
    Console.WriteLine(count);
  }
}
 
// This code is contributed by avanitrachhadiya2155
Output: 
3

 

Time Complexity: O(N), where N is the number of nodes in Tree.
 

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