bool IsValidBST(Node root, int min, int max) { if(root == null) return true; if (root.iData > min && root.iData < max && IsValidBST(root.LeftChild, min, root.iData) && IsValidBST(root.RightChild, root.iData, max)) return true; else return false; }

Not very difficult. Asked for answer in code. Interviewer took notes for each line I wrote. Was interested in each statement I made about the algorithm, esp. why I took each approach.

@Bala Recursively: Check node.left node.value. Also, that all subvalues of node.left node.value. Make sure you do checking for edge cases (ie. leaf nodes, etc).

recursive solution is a good option i guess.. its definitely easy to code. However i think we can have it done more efficiently if we just perform a BFS on the given tree. and every time just check tht the left child is smaller and right chiled is greater than or equal to the current node. If we successfully search the entire tree then it is a proper BST otherwise its not.

public static boolean checkTree(BSTNode node, Integer leftLimit, Integer rightLimit) { if(node == null) return true; if(leftLimit != null && leftLimit > node.info) { return false; } if(rightLimit != null && rightLimit < node.info) { return false; } return checkTree(node.left,leftLimit,node.info) && checkTree(node.right,node.info,rightLimit); }