Thursday Code Puzzler: Verification of a Binary Search Tree
Thursday is code puzzler day here at DZone. The idea is simple: solve the coding problem as efficiently as you can, in any language or framework that you find suitable.
Note: Even though there really is nothing stopping you from finding a solution to this on the internet, try to keep honest, and come up with your own answer. It's all about the participation!
Verify if the tree is a binary search treeDo you have code puzzlers that you'd like to share with the DZone community? If so, please submit here.
Given a tree structure, provide a method that will verify whether or not it is actually a binary search tree. The Tree consists of a number of Node objects, each of which has a value and a reference to a right and left Node object.
Note that a BST is different to a standard binary tree because of the following properties :
- The leftsubtreeof 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.
- There must be no duplicate nodes.
For more on Binary Search Trees, check this article on Wikipedia.
Catch up on all our previous puzzlers here.
Comments
Vijay Nathani replied on Thu, 2013/01/03 - 5:22am
Groovy code
@groovy.transform.Immutable class Node { int n; Node left, right; } class BinaryTree { def nodes = new HashSet() def count = 0 def verifyTreeStructure(Node root) { if (!root) return true count++; nodes.add(root.n) return (!root.left || root.left.n < root.n) && (!root.right || root.right.n > root.n) && verifyTreeStructure(root.left) && verifyTreeStructure(root.right) } static def verify(Node root) { def c = new BinaryTree() return c.verifyTreeStructure(root) && c.nodes.size() == c.count } } assert BinaryTree.verify(new Node(2,new Node(1,null,null),new Node(3,null,null)))Erik Colban replied on Thu, 2013/01/03 - 8:32pm
boolean verifyBST(Node root) { return verifyBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE); } boolean verifyBST(Node node, int min, int max) { if (node == null) return true; if (node.getKey() == min && node.getLeft() != null) return false; if (node.getKey() == max && node.getRight() != null) return false; return min <= node.getKey() && node.getKey() <= max && verifyBST(node.getLeft(), min, node.getKey() - 1) && verifyBST(node.getRight(), node.getKey() + 1, max); }Vijay Nathani replied on Fri, 2013/01/04 - 1:58am
in response to:
Erik Colban
This is a better solution than mine because my previous solution did not make certain checks. The solution in groovy is
@groovy.transform.Immutable class Node { int n; Node left, right; def boolean verify() { return verifyBST(Integer.MIN_VALUE, Integer.MAX_VALUE) } private def boolean verifyBST(long min, long max) { if (n < min || n > max) return false return (!left || left.verifyBST(min,n-1)) && (!right || right.verifyBST(n+1,max)) } } assert new Node(2,new Node(1,null,null),new Node(3,null,null)).verify()Erik Colban replied on Fri, 2013/01/04 - 4:09am
Vijay,
Are you assuming that the key is an int or a long? If the key is an int, then the arguments of verifyBST could be int's. The problem then is that if min == Integer.MIN_VALUE and n == min, then n - 1 == Integer.MAX_VALUE. I guess that is the reason you used long's.
I suggest the following change:
private def boolean verifyBST(int min, int max) { return min <= n && n <= max && (!left || min < n && left.verifyBST(min,n-1)) && (!right || n < max && right.verifyBST(n+1,max)) }