binary tree search

# Python3 Program for recursive binary search. 
  
# Returns index of x in arr if present, else -1 
def binarySearch (arr, l, r, x): 
  
    # Check base case 
    if r >= l: 
  
        mid = l + (r - l) // 2
  
        # If element is present at the middle itself 
        if arr[mid] == x: 
            return mid 
          
        # If element is smaller than mid, then it  
        # can only be present in left subarray 
        elif arr[mid] > x: 
            return binarySearch(arr, l, mid-1, x) 
  
        # Else the element can only be present  
        # in right subarray 
        else: 
            return binarySearch(arr, mid + 1, r, x) 
  
    else: 
        # Element is not present in the array 
        return -1
  
# Driver Code 
arr = [ 2, 3, 4, 10, 40 ] 
x = 10
  
# Function call 
result = binarySearch(arr, 0, len(arr)-1, x) 
  
if result != -1: 
    print ("Element is present at index % d" % result) 
else: 
    print ("Element is not present in array") 

/* This is just the seaching function you need to write the required code.
	Thank you. */

void searchNode(Node *root, int data)
{
    if(root == NULL)
    {
        cout << "Tree is empty\n";
        return;
    }

    queue<Node*> q;
    q.push(root);

    while(!q.empty())
    {
        Node *temp = q.front();
        q.pop();

        if(temp->data == data)
        {
            cout << "Node found\n";
            return;
        }

        if(temp->left != NULL)
            q.push(temp->left);
        if(temp->right != NULL)
            q.push(temp->right);
    }

    cout << "Node not found\n";
}

def binarySearch(arr, l, r, x): 
  
    while l <= r: 
  
        mid = l + (r - l) // 2; 
          
        # Check if x is present at mid 
        if arr[mid] == x: 
            return mid 
  
        # If x is greater, ignore left half 
        elif arr[mid] < x: 
            l = mid + 1
  
        # If x is smaller, ignore right half 
        else: 
            r = mid - 1
      
    # If we reach here, then the element 
    # was not present 
    return -1
  
# Driver Code 
arr = [ 2, 3, 4, 10, 40 ] 
x = 10
  
# Function call 
result = binarySearch(arr, 0, len(arr)-1, x) 
  
if result != -1: 
    print ("Element is present at index % d" % result) 
else: 
    print ("Element is not present in array")

import java.util.Scanner;

public class Binarysearch {

	public static void main(String[] args) {
		int[] x= {1,2,3,4,5,6,7,8,9,10,16,18,20,21};
		Scanner scan=new Scanner(System.in);
		System.out.println("enter the key:");
		int key=scan.nextInt();
		int flag=0;
		int low=0;
		int high=x.length-1;
		int mid=0;
		while(low<=high)
		{
			mid=(low+high)/2;
			if(key<x[mid])
			{
				high=mid-1;
			}
			else if(key>x[mid])
			{
				low=mid+1;
			}
			else if(key==x[mid])
			{
				flag++;
				System.out.println("found at index:"+mid);
				break;
			}
		}
		if(flag==0)
		{
			System.out.println("Not found");
		}
		

	}

}