binary-search-tree

This question already has an answer here: How do you validate a binary search tree? 30 answers Today I had an interview where I was asked to write a program which takes a Binary Tree and returns true if it is also a Binary Search Tree otherwise false. My Approach1: Perform an in-order traversal and store the elements in O(n) time. Now scan through the array/list of elements and check if element at i th index is greater than element at (i+1) th index. If such a condition is encountered, return false and break out of the loop. (This takes O(n) time). At the end return true. But this gentleman

I've been looking into this on Google and read the Collections entry in the SDK documentation, and turned up nothing. Is there a BST (any of its variants) implementation available out of the box with the iOS SDK? It seems odd that something so basic would be missing from a major development platform. Is their hash implementation just that magical? Or do the devs assume no one is going to do inserts/deletes on things that have an order? I can use NSSet for now, as I know most of us (myself included) aren't really writing anything with tons of computation on iOS that need a guaranteed access

How many binary search trees can be constructed from n distinct elements? And how can we find a mathematically proved formula for it? Example: If we have 3 distinct elements, say 1, 2, 3, there are 5 binary search trees. Given n elements, the number of binary search trees that can be made from those elements is given by the nth Catalan number (denoted C n ). This is equal to Intuitively, the Catalan numbers represent the number of ways that you can create a structure out of n elements that is made in the following way: Order the elements as 1, 2, 3, ..., n. Pick one of those elements to use as

What are the main differences between a Linked List and a BinarySearchTree? Is BST just a way of maintaining a LinkedList? My instructor talked about LinkedList and then BST but did't compare them or didn't say when to prefer one over another. This is probably a dumb question but I'm really confused. I would appreciate if someone can clarify this in a simple manner. Linked List: Item(1) -> Item(2) -> Item(3) -> Item(4) -> Item(5) -> Item(6) -> Item(7) Binary tree: Node(1) / Node(2) / \ / Node(3) RootNode(4) \ Node(5) \ / Node(6) \ Node(7) In a linked list, the items are linked together through

I already have implemented display of binary search tree . Here's the code , which paints the binary tree in a jpanel . public void paint(Graphics g) { super.paint(g); System.out.println(" in paint"); Graphics2D g2 = (Graphics2D) g; g2.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON); int num = bst.size; int y = 25; int nodes = 1; int level = 1; int length = getWidth(); Queue<Node> q = new LinkedList<Node>(); Queue<Integer> q2 = new LinkedList<Integer>(); q.add(bst.root); while (num > 0) { int pX = (int) Math.round(length / (2.0 * nodes)); int x = pX; for