Function to check whether a binary tree is binary search tree or not working

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Can someone tell me why is this not working? This seems correct to me

please someone look into this.

I am not able to find my mistake.

bool checkbst(node* root,int minValue,int maxValue)
{
   if(root==NULL)
   {

       return true;
   }
   else if(((root->data)>(minValue))&&
           ((root->data)>(maxValue))&&
           (checkbst(root->left,minValue,root->data))&&
           (checkbst(root->right,root->data,maxValue)))
   {

       return true;
   }
   else
   {

     return false;
   }
}

void isbst(node* root)
{
   if( checkbst(root,INT_MIN,INT_MAX))
   {
       cout<<"the tree is bst";
   }
}

You have a typo in checkbst, you are checking

((root->data)>(minValue))&&((root->data)>(maxValue))

while it probably should be

((root->data)>(minValue))&&((root->data)<(maxValue))

(notice the "less than" sign).

A program to check if a binary tree is BST or not, A binary search tree (BST) is a node based binary tree data structure which has It is assumed that you have helper functions minValue() and maxValue() that  A program to check if a binary tree is BST or not A binary search tree (BST) is a node based binary tree data structure which has the following properties. • The left subtree of a node contains only nodes with keys less than the node’s key.

Your code verifies that the keys are inside a range, but it does not verify if the children satisfy the bst condition respect to the root. That is, the keys in the left subtree must be lesser than the root and the keys in the right one greater. You should check if the children are not null before doing any comparison involving subtrees.

This version should work:

bool checkbst(node* root, int minValue,int maxValue)
{
  if (root == nullptr) 
    return true;

  if (not (root->data >= minValue && root->data <= maxvalue))
    return false;

  if (root->left)
    {
      if (root->data < root->left->data)
        if (not checkbst(root->left, minValue, maxValue))
          return false;
      else 
        return false;
    }

  // here the left subtree has been checked
  if (root->right)
    {
      if (root->data < root->right->data)
        return checkbst(root->right, minValue, maxValue);
      else
        return false;
    }

  return true; // everything is ok
}

Check if a Binary Tree is BST : Simple and Efficient Approach , A binary search tree (BST) is a node-based binary tree data structure which has the following properties. In this article, we will discuss a simple yet efficient approach to solve the above problem. Utility function to check if Binary Tree is BST. This article describes how to check whether a given tree is BST or not? This problem came in coding round of Microsoft. Problem Statement: Given a binary tree check whether it is a binary search tree or not. Solution. Algorithm:

I have checked your there is a small mistake code but there is a better way to do it. You just have to do the in order traversal of the given tree and store it in a array and then check if the elements in the array are sorted. If the elements are sorted then its a binary search tree else it will be a binary tree (which is a kind of basic difference between a binary tree and binary search tree).

There is a small mistake in your code

((root->data)>(maxValue))

should be

((root->data)<(maxValue))

Determine if given Binary Tree is a BST or not, Given a Binary Tree, determine if binary tree is a BST or not. the value at the left child and smaller than the value on the right child – does not work for all cases. Data structure to store a Binary Search Tree node. struct Node. {. int data;. Node *left, *right;. };. // Function to create a new binary tree node having given key. This can be solved in 2 ways: Method 1: perform a simple inorder traversal and keep the previous value of the node. If the current node is smaller than the previous node then it is not a binary search tree.

Here is a solution that is O(n) time complexity and O(1) space. It uses in-order tree traversal to confirm that the tree is sorted according to BST rules, but it does not rely on maintaining an auxiliary array of in-order traversed Nodes. However, because it does rely on recursion it's usage of the stack (i.e. stack depth) can reach O(logn).

struct Node
{
    int data;
    struct Node* left;
    struct Node* right;
};

bool isBSTHelper(Node* root, int& min, int& max)
{
    if (nullptr == root)
    {
        max = numeric_limits<int>::min(); // has meaning for LHS traversal.
        min = numeric_limits<int>::max(); // has meaning for RHS traversal.
        return true;
    }

    int lhsMax;
    int lhsMin;
    if (!isBSTHelper(root->left, lhsMin, lhsMax) ||
        lhsMax >= root->data)
    {
        return false;
    }

    int rhsMax;
    int rhsMin;
    if (!isBSTHelper(root->right, rhsMin, rhsMax) ||
        rhsMin <= root->data)
    {
        return false;
    }

    min = std::min(lhsMin, root->data);
    max = std::max(rhsMax, root->data);

    return true;
}

bool isBST(Node* root)
{
    int min;
    int max;
    return isBSTHelper(root, min, max);
}

Check if a binary tree is BST or not - Interview Problem, Problem Description: Given the root of a binary tree, check whether it is a binary Problem Note: The inorder traversal of a binary search tree explores tree The node structure for the BST passed to your function will be I attempted writing the following method which tells whether a Binary Tree is Binary Search Tree or not? I pass only half of the test cases. What am I doing wrong? boolean checkBST(Node root) {

Check if Binary Tree is Binary Search Tree, Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with  3) If a binary tree node has left and right sub-trees, then it is a part of a full binary tree by definition. In this case recursively check if the left and right sub-trees are also binary trees themselves. 4) In all other combinations of right and left sub-trees, the binary tree is not a full binary tree.

Check if a binary tree is binary search tree or not, You have a typo in checkbst, you are checking ((root->data)>(minValue))&&((root​->data)>(maxValue)). while it probably should be Objective: Given a Binary tree, find out whether its binary search tree or not. Input: A Binary Tree. Output: True or false based on whether tree is BST ot not. Approach: Method 1 : If tree doesn’t have duplicates. Do the inorder traversal of the given binary tree. check if the previously visited node is less than the current node value.

Validate Binary Search Tree, Problem Statement: Given a binary tree check whether it is a binary search tree or not. Solution. Algorithm: From the definition of BST, it may  Check if a binary tree is binary search tree or not we have written a program in C/C++ to verify whether a given binary tree is binary search tree or not. Check if Binary Tree is Binary

Comments
  • Edited to make the typo stand out better.
  • Your bst has only unique values?
  • yes it only has unique values.