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.
- 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.
From the above properties it naturally follows that:
- Each node (item in the tree) has a distinct key.
METHOD 1 (Simple but Wrong)
Following is a simple program. For each node, check if the left node of it is smaller than the node and right node of it is greater than the node.
C++
int isBST( struct node* node) { if (node == NULL) return 1; /* false if left is > than node */ if (node->left != NULL && node->left->data > node->data) return 0; /* false if right is < than node */ if (node->right != NULL && node->right->data < node->data) return 0; /* false if, recursively, the left or right is not a BST */ if (!isBST(node->left) || !isBST(node->right)) return 0; /* passing all that, it's a BST */ return 1; } // This code is contributed by shubhamsingh10 |
C
int isBST( struct node* node) { if (node == NULL) return 1; /* false if left is > than node */ if (node->left != NULL && node->left->data > node->data) return 0; /* false if right is < than node */ if (node->right != NULL && node->right->data < node->data) return 0; /* false if, recursively, the left or right is not a BST */ if (!isBST(node->left) || !isBST(node->right)) return 0; /* passing all that, it's a BST */ return 1; } |
Java
boolean isBST(Node node) { if (node == null ) return true ; /* False if left is > than node */ if (node.left != null && node.left.data > node.data) return false ; /* False if right is < than node */ if (node.right != null && node.right.data < node.data) return false ; /* False if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return false ; /* Passing all that, it's a BST */ return true ; } // This code is contributed by shubhamsingh10 |
Python3
def isBST(node): if (node = = None ): return 1 ''' false if left is > than node ''' if (node.left ! = None and node.left.data > node.data): return 0 ''' false if right is < than node ''' if (node.right ! = None and node.right.data < node.data): return 0 ''' false if, recursively, the left or right is not a BST ''' if (!isBST(node.left) or !isBST(node.right)): return 0 ''' passing all that, it's a BST ''' return 1 # This code is contributed by Shubham Singh |
C#
bool isBST(Node node) { if (node == null ) return true ; /* False if left is > than node */ if (node.left != null && node.left.data > node.data) return false ; /* False if right is < than node */ if (node.right != null && node.right.data < node.data) return false ; /* False if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return false ; /* Passing all that, it's a BST */ return true ; } // This code is contributed by Rajput-Ji |
Javascript
<script> function isBST(node) { if (node == null ) return true ; /* False if left is > than node */ if (node.left != null && node.left.data > node.data) return false ; /* False if right is < than node */ if (node.right != null && node.right.data < node.data) return false ; /* False if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return false ; /* Passing all that, it's a BST */ return true ; } // This code is contributed by avanitrachhadiya2155 </script> |
Time Complexity: O(n)
As we visit every node just once
Auxiliary Space: O(h)
Here h is the height of the tree and the extra space is used due to function call stack.
This approach is wrong as this will return true for below binary tree (and below tree is not a BST because 4 is in left subtree of 3)
METHOD 2 (Correct but not efficient)
For each node, check if max value in left subtree is smaller than the node and min value in right subtree greater than the node.
C++
/* Returns true if a binary tree is a binary search tree */ int isBST( struct node* node) { if (node == NULL) return 1; /* false if the max of the left is > than us */ if (node->left != NULL && maxValue(node->left) >= node->data) return 0; /* false if the min of the right is <= than us */ if (node->right != NULL && minValue(node->right) <= node->data) return 0; /* false if, recursively, the left or right is not a BST */ if (!isBST(node->left) || !isBST(node->right)) return 0; /* passing all that, it's a BST */ return 1; } // This code is contributed by shubhamsingh10 |
C
/* Returns true if a binary tree is a binary search tree */ int isBST( struct node* node) { if (node == NULL) return 1; /* false if the max of the left is > than us */ if (node->left!=NULL && maxValue(node->left) > node->data) return 0; /* false if the min of the right is <= than us */ if (node->right!=NULL && minValue(node->right) < node->data) return 0; /* false if, recursively, the left or right is not a BST */ if (!isBST(node->left) || !isBST(node->right)) return 0; /* passing all that, it's a BST */ return 1; } |
Java
/* Returns true if a binary tree is a binary search tree */ int isBST(Node node) { if (node == null ) return 1 ; /* false if the max of the left is > than us */ if (node.left != null && maxValue(node.left) >= node.data) return 0 ; /* false if the min of the right is <= than us */ if (node.right != null && minValue(node.right) <= node.data) return 0 ; /* false if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return 0 ; /* passing all that, it's a BST */ return 1 ; } // This code is contributed by akshitsaxenaa09. |
Python3
''' Returns true if a binary tree is a binary search tree ''' def isBST(node): if (node = = None ): return 1 ''' false if the max of the left is > than us ''' if (node.left ! = None and maxValue(node.left) > = node.data): return 0 ''' false if the min of the right is <= than us ''' if (node.right ! = None and minValue(node.right) < = node.data): return 0 ''' false if, recursively, the left or right is not a BST ''' if (!isBST(node.left) or !isBST(node.right)): return 0 ''' passing all that, it's a BST ''' return 1 # This code is contributed by Shubham Singh |
C#
/* Returns true if a binary tree is a binary search tree */ bool isBST(Node node) { if (node == null ) return true ; /* false if the max of the left is > than us */ if (node.left != null && maxValue(node.left) >= node.data) return false ; /* false if the min of the right is <= than us */ if (node.right != null && minValue(node.right) <= node.data) return false ; /* false if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return false ; /* passing all that, it's a BST */ return true ; } // This code is contributed by Shubham Singh |
Javascript
<script> function isBST(node) { if (node == null ) return true ; /* False if the max of the left is > than us */ if (node.left != null && maxValue(node.left) >= node.data) return false ; /* False if the min of the right is <= than us */ if (node.right != null && minValue(node.right) <= node.data) return false ; /* False if, recursively, the left or right is not a BST */ if (!isBST(node.left) || !isBST(node.right)) return false ; /* Passing all that, it's a BST */ return true ; } // This code is contributed by Shubham Singh </script> |
It is assumed that you have helper functions minValue() and maxValue() that return the min or max int value from a non-empty tree
Time Complexity: O(n^2)
As we visit every node just once and our helper method also takes O(n) time, so overall time complexity becomes O(n) * O(n) = O(n^2)
Auxiliary Space: O(h)
Here h is the height of the tree and the extra space is used due to function call stack.
METHOD 3 (Correct and Efficient):
Method 2 above runs slowly since it traverses over some parts of the tree many times. A better solution looks at each node only once. The trick is to write a utility helper function isBSTUtil(struct node* node, int min, int max) that traverses down the tree keeping track of the narrowing min and max allowed values as it goes, looking at each node only once. The initial values for min and max should be INT_MIN and INT_MAX — they narrow from there.
Note: This method is not applicable if there are duplicate elements with value INT_MIN or INT_MAX.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */ class node { public : int data; node* left; node* right; /* Constructor that allocates a new node with the given data and NULL left and right pointers. */ node( int data) { this ->data = data; this ->left = NULL; this ->right = NULL; } }; int isBSTUtil(node* node, int min, int max); /* Returns true if the given tree is a binary search tree (efficient version). */ int isBST(node* node) { return (isBSTUtil(node, INT_MIN, INT_MAX)); } /* Returns true if the given tree is a BST and its values are >= min and <= max. */ int isBSTUtil(node* node, int min, int max) { /* an empty tree is BST */ if (node==NULL) return 1; /* false if this node violates the min/max constraint */ if (node->data < min || node->data > max) return 0; /* otherwise check the subtrees recursively, tightening the min or max constraint */ return isBSTUtil(node->left, min, node->data-1) && // Allow only distinct values isBSTUtil(node->right, node->data+1, max); // Allow only distinct values } /* Driver code*/ int main() { node *root = new node(4); root->left = new node(2); root->right = new node(5); root->left->left = new node(1); root->left->right = new node(3); if (isBST(root)) cout<< "Is BST" ; else cout<< "Not a BST" ; return 0; } // This code is contributed by rathbhupendra |
C
#include <stdio.h> #include <stdlib.h> #include <limits.h> /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; int isBSTUtil( struct node* node, int min, int max); /* Returns true if the given tree is a binary search tree (efficient version). */ int isBST( struct node* node) { return (isBSTUtil(node, INT_MIN, INT_MAX)); } /* Returns true if the given tree is a BST and its values are >= min and <= max. */ int isBSTUtil( struct node* node, int min, int max) { /* an empty tree is BST */ if (node==NULL) return 1; /* false if this node violates the min/max constraint */ if (node->data < min || node->data > max) return 0; /* otherwise check the subtrees recursively, tightening the min or max constraint */ return isBSTUtil(node->left, min, node->data-1) && // Allow only distinct values isBSTUtil(node->right, node->data+1, max); // Allow only distinct values } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode( int data) { struct node* node = ( struct node*) malloc ( sizeof ( struct node)); node->data = data; node->left = NULL; node->right = NULL; return (node); } /* Driver program to test above functions*/ int main() { struct node *root = newNode(4); root->left = newNode(2); root->right = newNode(5); root->left->left = newNode(1); root->left->right = newNode(3); if (isBST(root)) printf ( "Is BST" ); else printf ( "Not a BST" ); getchar (); return 0; } |
Java
//Java implementation to check if given Binary tree //is a BST or not /* Class containing left and right child of current node and key value*/ class Node { int data; Node left, right; public Node( int item) { data = item; left = right = null ; } } public class BinaryTree { //Root of the Binary Tree Node root; /* can give min and max value according to your code or can write a function to find min and max value of tree. */ /* returns true if given search tree is binary search tree (efficient version) */ boolean isBST() { return isBSTUtil(root, Integer.MIN_VALUE, Integer.MAX_VALUE); } /* Returns true if the given tree is a BST and its values are >= min and <= max. */ boolean isBSTUtil(Node node, int min, int max) { /* an empty tree is BST */ if (node == null ) return true ; /* false if this node violates the min/max constraints */ if (node.data < min || node.data > max) return false ; /* otherwise check the subtrees recursively tightening the min/max constraints */ // Allow only distinct values return (isBSTUtil(node.left, min, node.data- 1 ) && isBSTUtil(node.right, node.data+ 1 , max)); } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node( 4 ); tree.root.left = new Node( 2 ); tree.root.right = new Node( 5 ); tree.root.left.left = new Node( 1 ); tree.root.left.right = new Node( 3 ); if (tree.isBST()) System.out.println( "IS BST" ); else System.out.println( "Not a BST" ); } } |
Python3
# Python program to check if a binary tree is bst or not INT_MAX = 4294967296 INT_MIN = - 4294967296 # A binary tree node class Node: # Constructor to create a new node def __init__( self , data): self .data = data self .left = None self .right = None # Returns true if the given tree is a binary search tree # (efficient version) def isBST(node): return (isBSTUtil(node, INT_MIN, INT_MAX)) # Retusn true if the given tree is a BST and its values # >= min and <= max def isBSTUtil(node, mini, maxi): # An empty tree is BST if node is None : return True # False if this node violates min/max constraint if node.data < mini or node.data > maxi: return False # Otherwise check the subtrees recursively # tightening the min or max constraint return (isBSTUtil(node.left, mini, node.data - 1 ) and isBSTUtil(node.right, node.data + 1 , maxi)) # Driver program to test above function root = Node( 4 ) root.left = Node( 2 ) root.right = Node( 5 ) root.left.left = Node( 1 ) root.left.right = Node( 3 ) if (isBST(root)): print ( "Is BST" ) else : print ( "Not a BST" ) # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System; // C# implementation to check if given Binary tree //is a BST or not /* Class containing left and right child of current node and key value*/ public class Node { public int data; public Node left, right; public Node( int item) { data = item; left = right = null ; } } public class BinaryTree { //Root of the Binary Tree public Node root; /* can give min and max value according to your code or can write a function to find min and max value of tree. */ /* returns true if given search tree is binary search tree (efficient version) */ public virtual bool BST { get { return isBSTUtil(root, int .MinValue, int .MaxValue); } } /* Returns true if the given tree is a BST and its values are >= min and <= max. */ public virtual bool isBSTUtil(Node node, int min, int max) { /* an empty tree is BST */ if (node == null ) { return true ; } /* false if this node violates the min/max constraints */ if (node.data < min || node.data > max) { return false ; } /* otherwise check the subtrees recursively tightening the min/max constraints */ // Allow only distinct values return (isBSTUtil(node.left, min, node.data - 1) && isBSTUtil(node.right, node.data + 1, max)); } /* Driver program to test above functions */ public static void Main( string [] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(4); tree.root.left = new Node(2); tree.root.right = new Node(5); tree.root.left.left = new Node(1); tree.root.left.right = new Node(3); if (tree.BST) { Console.WriteLine( "IS BST" ); } else { Console.WriteLine( "Not a BST" ); } } } // This code is contributed by Shrikant13 |
Javascript
<script> // Javascript implementation to // check if given Binary tree // is a BST or not /* Class containing left and right child of current node and key value*/ class Node { constructor(item) { this .data=item; this .left= this .right= null ; } } //Root of the Binary Tree let root; /* can give min and max value according to your code or can write a function to find min and max value of tree. */ /* returns true if given search tree is binary search tree (efficient version) */ function isBST() { return isBSTUtil(root, Number.MIN_VALUE, Number.MAX_VALUE); } /* Returns true if the given tree is a BST and its values are >= min and <= max. */ function isBSTUtil(node,min,max) { /* an empty tree is BST */ if (node == null ) return true ; /* false if this node violates the min/max constraints */ if (node.data < min || node.data > max) return false ; /* otherwise check the subtrees recursively tightening the min/max constraints */ // Allow only distinct values return (isBSTUtil(node.left, min, node.data-1) && isBSTUtil(node.right, node.data+1, max)); } /* Driver program to test above functions */ root = new Node(4); root.left = new Node(2); root.right = new Node(5); root.left.left = new Node(1); root.left.right = new Node(3); if (isBST()) document.write( "IS BST<br>" ); else document.write( "Not a BST<br>" ); // This code is contributed by rag2127 </script> |
Output:
IS BST
Time Complexity: O(n)
Auxiliary Space: O(1) if Function Call Stack size is not considered, otherwise O(h) where h is height of the tree
Simplified Method 3
We can simplify method 2 using NULL pointers instead of INT_MIN and INT_MAX values.
C++
// C++ program to check if a given tree is BST. #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */ struct Node { int data; struct Node* left, *right; }; // Returns true if given tree is BST. bool isBST(Node* root, Node* l=NULL, Node* r=NULL) { // Base condition if (root == NULL) return true ; // if left node exist then check it has // correct data or not i.e. left node's data // should be less than root's data if (l != NULL and root->data <= l->data) return false ; // if right node exist then check it has // correct data or not i.e. right node's data // should be greater than root's data if (r != NULL and root->data >= r->data) return false ; // check recursively for every node. return isBST(root->left, l, root) and isBST(root->right, root, r); } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct Node* newNode( int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } /* Driver program to test above functions*/ int main() { struct Node *root = newNode(3); root->left = newNode(2); root->right = newNode(5); root->left->left = newNode(1); root->left->right = newNode(4); if (isBST(root,NULL,NULL)) cout << "Is BST" ; else cout << "Not a BST" ; return 0; } |
Java
// Java program to check if a given tree is BST. class Sol { // A binary tree node has data, pointer to //left child && a pointer to right child / static class Node { int data; Node left, right; }; // Returns true if given tree is BST. static boolean isBST(Node root, Node l, Node r) { // Base condition if (root == null ) return true ; // if left node exist then check it has // correct data or not i.e. left node's data // should be less than root's data if (l != null && root.data <= l.data) return false ; // if right node exist then check it has // correct data or not i.e. right node's data // should be greater than root's data if (r != null && root.data >= r.data) return false ; // check recursively for every node. return isBST(root.left, l, root) && isBST(root.right, root, r); } // Helper function that allocates a new node with the //given data && null left && right pointers. / static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Driver code public static void main(String args[]) { Node root = newNode( 3 ); root.left = newNode( 2 ); root.right = newNode( 5 ); root.left.left = newNode( 1 ); root.left.right = newNode( 4 ); if (isBST(root, null , null )) System.out.print( "Is BST" ); else System.out.print( "Not a BST" ); } } // This code is contributed by Arnab Kundu |
Python3
""" Program to check if a given Binary Tree is balanced like a Red-Black Tree """ # Helper function that allocates a new # node with the given data and None # left and right poers. class newNode: # Construct to create a new node def __init__( self , key): self .data = key self .left = None self .right = None # Returns true if given tree is BST. def isBST(root, l = None , r = None ): # Base condition if (root = = None ) : return True # if left node exist then check it has # correct data or not i.e. left node's data # should be less than root's data if (l ! = None and root.data < = l.data) : return False # if right node exist then check it has # correct data or not i.e. right node's data # should be greater than root's data if (r ! = None and root.data > = r.data) : return False # check recursively for every node. return isBST(root.left, l, root) and \ isBST(root.right, root, r) # Driver Code if __name__ = = '__main__' : root = newNode( 3 ) root.left = newNode( 2 ) root.right = newNode( 5 ) root.right.left = newNode( 1 ) root.right.right = newNode( 4 ) #root.right.left.left = newNode(40) if (isBST(root, None , None )): print ( "Is BST" ) else : print ( "Not a BST" ) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to check if a given tree is BST. using System; class GFG { // A binary tree node has data, pointer to //left child && a pointer to right child / public class Node { public int data; public Node left, right; }; // Returns true if given tree is BST. static Boolean isBST(Node root, Node l, Node r) { // Base condition if (root == null ) return true ; // if left node exist then check it has // correct data or not i.e. left node's data // should be less than root's data if (l != null && root.data <= l.data) return false ; // if right node exist then check it has // correct data or not i.e. right node's data // should be greater than root's data if (r != null && root.data >= r.data) return false ; // check recursively for every node. return isBST(root.left, l, root) && isBST(root.right, root, r); } // Helper function that allocates a new node with the //given data && null left && right pointers. / static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = node.right = null ; return (node); } // Driver code public static void Main(String []args) { Node root = newNode(3); root.left = newNode(2); root.right = newNode(5); root.left.left = newNode(1); root.left.right = newNode(4); if (isBST(root, null , null )) Console.Write( "Is BST" ); else Console.Write( "Not a BST" ); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to check if a given tree is BST. class Node { constructor(data) { this .left = null ; this .right = null ; this .data = data; } } // Returns true if given tree is BST. function isBST(root, l, r) { // Base condition if (root == null ) return true ; // if left node exist then check it has // correct data or not i.e. left node's data // should be less than root's data if (l != null && root.data <= l.data) return false ; // if right node exist then check it has // correct data or not i.e. right node's data // should be greater than root's data if (r != null && root.data >= r.data) return false ; // check recursively for every node. return isBST(root.left, l, root) && isBST(root.right, root, r); } // Helper function that allocates a new node with the //given data && null left && right pointers. / function newNode(data) { let node = new Node(data); return (node); } let root = newNode(3); root.left = newNode(2); root.right = newNode(5); root.left.left = newNode(1); root.left.right = newNode(4); if (isBST(root, null , null )) document.write( "Is BST" ); else document.write( "Not a BST" ); </script> |
Output:
Not a BST
Time Complexity: O(n)
As we visit every node just once
Auxiliary Space: O(h)
Here h is the height of the tree and the extra space is used due to function call stack.
Thanks to Abhinesh Garhwal for suggesting above solution.
METHOD 4(Using In-Order Traversal)
Thanks to LJW489 for suggesting this method.
1) Do In-Order Traversal of the given tree and store the result in a temp array.
2) This method assumes that there are no duplicate values in the tree
3) Check if the temp array is sorted in ascending order, if it is, then the tree is BST.
Time Complexity: O(n)
We can avoid the use of a Auxiliary Array. While doing In-Order traversal, we can keep track of previously visited node. If the value of the currently visited node is less than the previous value, then tree is not BST. Thanks to ygos for this space optimization.
C++
bool isBST(node* root) { static node *prev = NULL; // traverse the tree in inorder fashion // and keep track of prev node if (root) { if (!isBST(root->left)) return false ; // Allows only distinct valued nodes if (prev != NULL && root->data <= prev->data) return false ; prev = root; return isBST(root->right); } return true ; } // This code is contributed by rathbhupendra |
C
bool isBST( struct node* root) { static struct node *prev = NULL; // traverse the tree in inorder fashion and keep track of prev node if (root) { if (!isBST(root->left)) return false ; // Allows only distinct valued nodes if (prev != NULL && root->data <= prev->data) return false ; prev = root; return isBST(root->right); } return true ; } |
Java
// Java implementation to check if given Binary tree // is a BST or not /* Class containing left and right child of current node and key value*/ class Node { int data; Node left, right; public Node( int item) { data = item; left = right = null ; } } public class BinaryTree { // Root of the Binary Tree Node root; // To keep tract of previous node in Inorder Traversal Node prev; boolean isBST() { prev = null ; return isBST(root); } /* Returns true if given search tree is binary search tree (efficient version) */ boolean isBST(Node node) { // traverse the tree in inorder fashion and // keep a track of previous node if (node != null ) { if (!isBST(node.left)) return false ; // allows only distinct values node if (prev != null && node.data <= prev.data ) return false ; prev = node; return isBST(node.right); } return true ; } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node( 4 ); tree.root.left = new Node( 2 ); tree.root.right = new Node( 5 ); tree.root.left.left = new Node( 1 ); tree.root.left.right = new Node( 3 ); if (tree.isBST()) System.out.println( "IS BST" ); else System.out.println( "Not a BST" ); } } |
Python3
# Python implementation to check if # given Binary tree is a BST or not # A binary tree node containing data # field, left and right pointers class Node: # constructor to create new node def __init__( self , val): self .data = val self .left = None self .right = None # global variable prev - to keep track # of previous node during Inorder # traversal prev = None # function to check if given binary # tree is BST def isbst(root): # prev is a global variable global prev prev = None return isbst_rec(root) # Helper function to test if binary # tree is BST # Traverse the tree in inorder fashion # and keep track of previous node # return true if tree is Binary # search tree otherwise false def isbst_rec(root): # prev is a global variable global prev # if tree is empty return true if root is None : return True if isbst_rec(root.left) is False : return False # if previous node'data is found # greater than the current node's # data return false if prev is not None and prev.data > root.data: return False # store the current node in prev prev = root return isbst_rec(root.right) # driver code to test above function root = Node( 4 ) root.left = Node( 2 ) root.right = Node( 5 ) root.left.left = Node( 1 ) root.left.right = Node( 3 ) if isbst(root): print ( "is BST" ) else : print ( "not a BST" ) # This code is contributed by # Shweta Singh(shweta44) |
C#
// C# implementation to check if // given Binary tree is a BST or not using System; /* Class containing left and right child of current node and key value*/ class Node { public int data; public Node left, right; public Node( int item) { data = item; left = right = null ; } } public class BinaryTree { // Root of the Binary Tree Node root; // To keep tract of previous node // in Inorder Traversal Node prev; Boolean isBST() { prev = null ; return isBST(root); } /* Returns true if given search tree is binary search tree (efficient version) */ Boolean isBST(Node node) { // traverse the tree in inorder fashion and // keep a track of previous node if (node != null ) { if (!isBST(node.left)) return false ; // allows only distinct values node if (prev != null && node.data <= prev.data ) return false ; prev = node; return isBST(node.right); } return true ; } // Driver Code public static void Main(String []args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(4); tree.root.left = new Node(2); tree.root.right = new Node(5); tree.root.left.left = new Node(1); tree.root.left.right = new Node(3); if (tree.isBST()) Console.WriteLine( "IS BST" ); else Console.WriteLine( "Not a BST" ); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation to check if given Binary tree // is a BST or not /* Class containing left and right child of current node and key value*/ class Node { constructor(item) { this .data = item; this .left = this .right= null ; } } // Root of the Binary Tree let root; // To keep tract of previous node in Inorder Traversal let prev; function isBST() { prev = null ; return _isBST(root); } /* Returns true if given search tree is binary search tree (efficient version) */ function _isBST(node) { // traverse the tree in inorder fashion and // keep a track of previous node if (node != null ) { if (!_isBST(node.left)) return false ; // allows only distinct values node if (prev != null && node.data <= prev.data ) return false ; prev = node; return _isBST(node.right); } return true ; } /* Driver program to test above functions */ root = new Node(4); root.left = new Node(2); root.right = new Node(5); root.left.left = new Node(1); root.left.right = new Node(3); if (isBST()) document.write( "IS BST" ); else document.write( "Not a BST" ); // This code is contributed by unknown2108 </script> |
The use of a static variable can also be avoided by using a reference to the prev node as a parameter.
C++
// C++ program to check if a given tree is BST. #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */ struct Node { int data; struct Node* left, *right; Node( int data) { this ->data = data; left = right = NULL; } }; bool isBSTUtil( struct Node* root, Node *&prev) { // traverse the tree in inorder fashion and // keep track of prev node if (root) { if (!isBSTUtil(root->left, prev)) return false ; // Allows only distinct valued nodes if (prev != NULL && root->data <= prev->data) return false ; prev = root; return isBSTUtil(root->right, prev); } return true ; } bool isBST(Node *root) { Node *prev = NULL; return isBSTUtil(root, prev); } /* Driver program to test above functions*/ int main() { struct Node *root = new Node(3); root->left = new Node(2); root->right = new Node(5); root->left->left = new Node(1); root->left->right = new Node(4); if (isBST(root)) cout << "Is BST" ; else cout << "Not a BST" ; return 0; } |
Java
// Java program to check if a given tree is BST. import java.io.*; class GFG { /* A binary tree node has data, pointer to left child and a pointer to right child */ public static class Node { public int data; public Node left, right; public Node( int data) { this .data = data; left = right = null ; } }; static Node prev; static Boolean isBSTUtil(Node root) { // traverse the tree in inorder fashion and // keep track of prev node if (root != null ) { if (!isBSTUtil(root.left)) return false ; // Allows only distinct valued nodes if (prev != null && root.data <= prev.data) return false ; prev = root; return isBSTUtil(root.right); } return true ; } static Boolean isBST(Node root) { return isBSTUtil(root); } // Driver Code public static void main (String[] args) { Node root = new Node( 3 ); root.left = new Node( 2 ); root.right = new Node( 5 ); root.left.left = new Node( 1 ); root.left.right = new Node( 4 ); if (isBST(root)) System.out.println( "Is BST" ); else System.out.println( "Not a BST" ); } } // This code is contributed by Shubham Singh |
Python3
# Python3 program to check # if a given tree is BST. import math # A binary tree node has data, # pointer to left child and # a pointer to right child class Node: def __init__( self , data): self .data = data self .left = None self .right = None def isBSTUtil(root, prev): # traverse the tree in inorder fashion # and keep track of prev node if (root ! = None ): if (isBSTUtil(root.left, prev) = = True ): return False # Allows only distinct valued nodes if (prev ! = None and root.data < = prev.data): return False prev = root return isBSTUtil(root.right, prev) return True def isBST(root): prev = None return isBSTUtil(root, prev) # Driver Code if __name__ = = '__main__' : root = Node( 3 ) root.left = Node( 2 ) root.right = Node( 5 ) root.right.left = Node( 1 ) root.right.right = Node( 4 ) #root.right.left.left = Node(40) if (isBST(root) = = None ): print ( "Is BST" ) else : print ( "Not a BST" ) # This code is contributed by Srathore |
C#
// C# program to check if a given tree is BST. using System; public class GFG { /* A binary tree node has data, pointer to left child and a pointer to right child */ public class Node { public int data; public Node left, right; public Node( int data) { this .data = data; left = right = null ; } }; static Node prev; static Boolean isBSTUtil(Node root) { // traverse the tree in inorder fashion and // keep track of prev node if (root != null ) { if (!isBSTUtil(root.left)) return false ; // Allows only distinct valued nodes if (prev != null && root.data <= prev.data) return false ; prev = root; return isBSTUtil(root.right); } return true ; } static Boolean isBST(Node root) { return isBSTUtil(root); } // Driver Code public static void Main(String[] args) { Node root = new Node(3); root.left = new Node(2); root.right = new Node(5); root.left.left = new Node(1); root.left.right = new Node(4); if (isBST(root)) Console.WriteLine( "Is BST" ); else Console.WriteLine( "Not a BST" ); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to check if a given tree is BST. class Node { constructor(data) { this .left = null ; this .right = null ; this .data = data; } } let prev; function isBSTUtil(root) { // traverse the tree in inorder fashion and // keep track of prev node if (root != null ) { if (!isBSTUtil(root.left)) return false ; // Allows only distinct valued nodes if (prev != null && root.data <= prev.data) return false ; prev = root; return isBSTUtil(root.right); } return true ; } function isBST(root) { return isBSTUtil(root); } let root = new Node(3); root.left = new Node(2); root.right = new Node(5); root.left.left = new Node(1); root.left.right = new Node(4); if (isBST(root)) document.write( "Is BST" ); else document.write( "Not a BST" ); // This code is contributed by divyeshrabadiya07. </script> |
Output:
Not a BST
Time Complexity: O(n)
As we visit every node just once
Auxiliary Space: O(h)
Here h is the height of the tree and the extra space is used due to function call stack.
Sources:
http://en.wikipedia.org/wiki/Binary_search_tree
http://cslibrary.stanford.edu/110/BinaryTrees.html
Please write comments if you find any bug in the above programs/algorithms or other ways to solve the same problem.