CHECKBST.java

/*Given the root of a binary tree. Check whether it is a BST or not.
Note: We are considering that BSTs can not contain duplicate Nodes.
A BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Examples:

Input:
   2
 /    \
1      3
        \
         5
Output: true 
Explanation: 
The left subtree of every node contains smaller keys and right subtree of every node contains greater. Hence, the tree is a BST.
Input:
  2
   \
    7
     \
      6
       \
        9
Output: false 
Explanation: 
Since the node with value 7 has right subtree nodes with keys less than 7, this is not a BST. 
Input:
   10
 /    \
5      20
      /   \
     9     25
Output: false
Explanation: The node is present in the right subtree of 10, but it is smaller than 10.
Expected Time Complexity: O(n) 
Expected Auxiliary Space: O(Height of the tree)
where n is the number of nodes in the given tree

Constraints:
1 ≤ Number of nodes ≤ 105
1 ≤ Data of a node ≤ 105 */

class Solution {
    // Function to check whether a Binary Tree is BST or not.
    boolean isBST(Node root) {
        ArrayList<Integer>node=new ArrayList<>();

        check(root,node);
        
        for(int i=0;i<node.size()-1;i++)
        {
            if(node.get(i) >= node.get(i+1))
            {
                return false;
            }
        }
        return true;
    }
    void check(Node root,ArrayList<Integer>node)
    {
        if(root==null)
        {
            return;
        }
        check(root.left,node);
        node.add(root.data);
        check(root.right,node);
    }
    
}