valid BST -- Leetcode

From Geeksforgeeks:

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.

/* Returns true if a binary tree is a binary search tree */ int isBST(struct 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); }

It is assumed that you have helper functions minValue() and maxValue() that return the min or max int value from a non-empty tree

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.

/* 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) 

Implementation:

#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 }   Time Complexity: O(n) Auxiliary Space : O(1) if Function Call Stack size is not considered, otherwise O(n) 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. 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 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. 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; } The use of static variable can also be avoided by using reference to prev node as a parameter (Similar to this post).