**Is This a Binary Search Tree?**

### Problem Statement :

For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a binary search tree? Complete the function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must return a boolean denoting whether or not the binary tree is a binary search tree. You may have to write one or more helper functions to complete this challenge. Input Format You are not responsible for reading any input from stdin. Hidden code stubs will assemble a binary tree and pass its root node to your function as an argument. Constraints 0 <= data <= 10^4 Output Format You are not responsible for printing any output to stdout. Your function must return true if the tree is a binary search tree; otherwise, it must return false. Hidden code stubs will print this result as a Yes or No answer on a new line.

### Solution :

` ````
Solution in C :
In C ++ :
The Node struct is defined as follows:
struct Node {
int data;
Node* left;
Node* right;
}
*/ enum comp {LESS, GREATER};
bool checkBST(Node* root,int minVal, int maxVal ){
if(root==0)
return true;
int nVal=root->data;
if(nVal<=minVal || nVal>=maxVal)
return false;
return checkBST(root->left,minVal,nVal) && checkBST(root->right,nVal, maxVal);
}
bool checkBST(Node* root) {
if(root==0)
return true;
return checkBST(root->left, -1, root->data) && checkBST(root->right,root->data, 10001);
}
In Java :
The Node class is defined as follows:
class Node {
int data;
Node left;
Node right;
}
*/
private boolean checkBST(Node n, int min, int max) {
if(n == null) return true;
return n.data > min && n.data < max && checkBST(n.left, min, n.data) && checkBST(n.right, n.data, max);
}
boolean checkBST(Node root) {
if(root == null) return true;
if(count == 0) return true;
return checkBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
In python3 :
from collections import deque
""" Node is defined as
class node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
"""
def check_binary_search_tree_(root, lowest_value=0, highest_value=10000):
min_v = lowest_value - 1
max_v = highest_value + 1
q = deque([(root, min_v, max_v)])
while q:
node, min_val, max_val = q.popleft()
if not node: continue
if node.data >= max_val or node.data <= min_val: return False
if node.left: q.append((node.left, min_val, node.data))
if node.right: q.append((node.right, node.data, max_val))
return True
```

## View More Similar Problems

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →## Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →## Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →## Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →