Check if a Binary Tree is a Binary Search Tree

So to repeat the question here: given a binary tree, check if it is a Binary Search Tree.

To solve this problem, the first thing that would probably come to our mind is to recursively check each node and compare its value and the value of its left child and that of its right child, to make sure left child’s value ≤ node’s value ≤ right child’s value. This is a good start, but it’s not completely right, because in a BST not only the direct children’s values must satisfy the above condition, but also all descendants from the left child must be smaller than or equal to the node’s value, and all descendants from right child must be greater than or equal to it. For example, if the root is 20, its left child is 10, and the left child has its own left child 5 and right child 15, the above condition is satisfied. However, if the left child’s right child has a value 25, this cannot be a BST since 25 is larger than the root value 20.

So how do we make sure that all the descendants on the left and the right side of a root node all satisfies the definition of BST? The In-Order traversal came to our rescue. By definition the In-Order traversal on a BST will visit each node in the correct ascending order. We can thus use such a traversing method, just to modify it a little bit so the recursive calls carry an extra inout parameter that remembers the last visited node’s value.

Here is the code that implements this method, with three test cases of good and bad BSTs included at the end:

and voila, the code does correctly identified true and fake BSTs.

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