binary-tree

I need an IntervalTree or RangeTree implementation in Java, and am having trouble finding one with working deletion support. There's a built-in one at sun.jvm.hotspot.utilities.IntervalTree , but the deleteNode method in the RBTree superclass states: /** * FIXME: this does not work properly yet for augmented red-black * trees since it doesn't update nodes. Need to figure out exactly * from which points we need to propagate updates upwards. */ Trying to delete nodes from a tree ends up throwing the exception: Node's max endpoint was not updated properly How difficult would it be to properly

Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i

I have a tree data type: data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a ...and I need to make it an instance of Show , without using deriving . I have found that nicely displaying a little branch with two leaves is easy: instance (Show a, Show b) => Show (Tree a b) where show (Leaf x) = show x show (Branch val l r) = " " ++ show val ++ "\n" ++ show l ++ " " ++ show r But how can I extend a nice structure to a tree of arbitrary size? It seems like determining the spacing would require me to know just how many leaves will be at the very bottom (or maybe just how many leaves there are in

How to find the Nth largest node in a BST? Do I keep a count variable while doing In Order Traversal of a BST? Return the element when the count = N??? IVlad See my answer here . You can do this in O(log n) on average where n = number of nodes. Worst case is still O(n) IF the tree isn't balanced (always O(log n) if it is balanced however). In order traversal is always O(n) however. Vallabh Patade The idea is very simple: traverse the tree in decreasing order of the values of each node. When you reach the Nth node, print that node value. Here is the recursive code. void printNthNode(Node* root,