Trees: Is This a Binary Search Tree?


Problem Statement :


For the purposes of this challenge, we define a binary search tree to be a binary tree with the following properties:

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.
The data value of every node is distinct.
For example, the image on the left below is a valid BST. The one on the right fails on several counts:
- All of the numbers on the right branch from the root are not larger than the root.
- All of the numbers on the right branch from node 5 are not larger than 5.
- All of the numbers on the left branch from node 5 are not smaller than 5.
- The data value 1 is repeated.


Function Description

Complete the function checkBST in the editor below. It must return a boolean denoting whether or not the binary tree is a binary search tree.

checkBST has the following parameter(s):

root: a reference to the root node of a tree to test
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

Your function must return a boolean true if the tree is a binary search tree. Otherwise, it must return false.


Solution :



title-img 




                        Solution in C++ :

In   C++ :






   struct Node {
      int data;
      Node* left;
      Node* right;
   }
*/
bool soy=true;
int mini(int a, int b)
    {
    return a<b ? a:b;
}
int maxi(int a, int b)
    {
    return a>b ? a:b;
}
typedef pair<int, int> ii;
#define f first
#define s second
ii revisar(Node* root)
{
    ii h(-1, 10005), iz(10005,10005), de(-1,-1);
     if (root->left!=NULL)
     {
         iz=revisar(root->left);
         if (iz.s >= root->data or iz.f >= root->data) soy=false;
     }
    if (root->right!=NULL)
    {
        de=revisar(root->right);    
        if (de.f <= root->data or de.s <= root->data) soy=false;
    }
    
    
    return ii{mini(root->data, iz.f),maxi(root->data, de.s) };
}


   bool checkBST(Node* root) {
      soy=true;
       if (root!=NULL)
        revisar(root);
       //else return false;
       return soy;
   }
                    


                        Solution in Java :

In   Java :






    class Node {
        int data;
        Node left;
        Node right;
     }
*/
    boolean checkBST(Node root) {  
        //return fasle;
        return checkBSTHelper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
     
    
    private boolean checkBSTHelper(Node n, int min, int max) {
        if (n == null) return true;
        if (n.data <= min || n.data >= max) return false;
        return checkBSTHelper(n.left, min, n.data) && checkBSTHelper(n.right, n.data, max);
    }
                    


                        Solution in Python : 
                            
In   Python3 :






""" Node is defined as
class node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
"""

def check(node, max_val = float('inf'), min_val = float('-inf')):
    if not node:
        return True
    if node.data <= min_val or node.data >= max_val:
        return False
    return check(node.left, node.data, min_val) and check(node.right, max_val, node.data)

def check_binary_search_tree_(root):
    return check(root)


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