C++ Program to Check if a Binary Tree is a BST

Binary Search Tree is a binary tree data structure in which we have 3 properties −

• The left subtree of a binary search tree of a node contains only nodes with keys lesser than the node’s key.

• The right subtree of a binary search tree node contains only nodes with keys greater than the node’s key.

• The left and right of a subtree each must also be a binary search tree.

Algorithm

Begin
   <strong>function BSTUtill()</strong>
      If node is equals to NULL then
         Return 1.
      If data of node is less than minimum or greater than
      maximum data then
         Return 0.
      Traverse left and right sub-trees recursively. 
End.

Example Code

#include <iostream>
#include <cstdlib>
#include <climits>
using namespace std;
struct n {
   int d;
   n* l;
   n* r;
};
int BSTUtil(n* node, int min, int max);
int isBST(n* node) {
   return(BSTUtil(node, INT_MIN, INT_MAX));
}
int BSTUtil(struct n* node, int min, int max) {
   if (node==NULL)
      return 1;
   if (node->d < min || node->d > max)
      return 0;
      return BSTUtil(node->l, min, node->d - 1) && BSTUtil(node->r, node->d + 1, max);
}
n* newN(int d) {
   n* nod = new n;
   nod->d = d;
   nod->l = NULL;
   nod->r = NULL;
   return nod;
}
int main() {
   n *root = newN(7);
   root->l = newN(6);
   root->r = newN(10);
   root->l->l = newN(2);
   root->l->r = newN(4);
   if (isBST(root))
      cout<<"The Given Binary Tree is a BST"<<endl;
   else
      cout<<"The Given Binary Tree is not a BST"<<endl;
      n *root1 = newN(10);
      root1->l = newN(6);
      root1->r = newN(11);
      root1->l->l = newN(2);
      root1->l->r = newN(7);
      if (isBST(root1))
         cout<<"The Given Binary Tree is a BST"<<endl;
      else
         cout<<"The Given Binary Tree is not a BST"<<endl;
      return 0;
}

Output

The Given Binary Tree is not a BST
The Given Binary Tree is a BST