binary-search-tree

问题 Ok so I thought it was fixed, but I'm getting totally inconsistent results. I rewrote it kind of from scratch to start fresh and here are my results. I get no errors, no crashing, it just doesn't remove them. It just totally messes up the tree and gives me a ton more leaves, and mixes everything up. Not sure where else to go template <class T> void BST<T>::remove(struct Node<T>*& root, const T& x) { Node<T>* ptr = root; bool found = false; Node<T>* parent; while (ptr != NULL && !found) { if

问题 I'm trying to add strings to a Binary Search Tree using a recursive insert method (the usual for BSTs, IIRC) so I can later print them out using recursion as well. Trouble is, I've been getting a segmentation faults I don't really understand. Related code follows (this block of code is from my main function): #include <stdio.h> #include <stdlib.h> #include <string.h> #include <ctype.h> // Stores the size of the C-strings we will use; // Standardized to 100 (assignment specifications say //

问题 Hi I am trying to print out level order of a binary search tree in the format (node element, parent node element) . I am currently using queues to get the level order but I am having a hard time getting the parent node. Is it possible to do with queues? If so how would I go about doing it? If not what is a more optimal way of doing it? Thank you! For example with the following tree: 6 / \ 5 7 Level 0: (6, null) Level 1: (5, 6) (7, 6) 回答1: So unfortunately there isn't a way of doing this

问题 I'm working with binary tree in python. I need to create a method which searches the tree and return the best node where a new value can be inserted. But i'm have trouble returning a value from this recursive function. I'm a total newbie in python. def return_key(self, val, node): if(val < node.v): if(node.l != None): self.return_key(val, node.l) else: print node.v return node else: if(node.r != None): #print node.v self.return_key(val, node.r) else: print node.v return node Printing node.v

问题 class BinaryNode: def __init__(self, value): self.data = value self.left = None self.right = None def contains(root, value): if root is None: return False if value == root.data: return True if value < root.data: return contains(root.left, value) else: return contains(root.right, value) def insert(root, value): if root is None: root = BinaryNode(value) else: if value > root.data: if root.right is None: root.right = BinaryNode(value) else: return insert(root.right, value) else: if root.left is

问题 I have postorder array of a BST tree size n how do i show there is only one BST that can be constructed form it. I know I can rebuild the tree if I add nodes from right to left but how do I show there is only one right tree? I have tried saying there are two possible trees and tried showing it is not possible but got stuck 回答1: It is possible only because it is a BST. Recall that for a Binary tree to be a valid Binary Search Tree: -Left subtrees' values must be less than root's value -Right

问题 There is a plenty of solutions how to find closest lower and upper keys in binary tree in imperative languages, but a lack of same questions for doing it in purely functional style like that of Haskell. I'm curious to learn how it's possible to walk around a binary serch tree ahead of meeting both closest keys. There is a function and some pattern matches I've so far: data TreeMap v = Leaf | Node { pair::(Integer, v), l::TreeMap v, r::TreeMap v} deriving (Show, Read, Eq, Ord) closestLess ::

问题 I'm trying to write a function to remove a node from a binary tree. I haven't coded the function yet, and I am trying to think about the different conditions I should consider for removing a node. I am guessing that the possible conditions are: The node has no children The node has one child The node has 2 children In each of these cases what would be the algorithm to perform a delete function? 回答1: This is something you would find in any standard textbook about algorithms, but let's suppose

问题 I'm trying to follow the BST algorithm in "Data Structures and Algorithms" by Granville Barnett, but I don't understand the node-deletion algorithm it describes below. Section 3.3 (p. 22) Removing a node from a BST is fairly straightforward, with four cases to consider: the value to remove is a leaf node; or the value to remove has a right subtree, but no left subtree; or the value to remove has a left subtree, but no right subtree; or the value to remove has both a left and right subtree in

问题 One of the answers in our powerpoint says it is n/2 leaves, but I am seeing another answer which says (n+1)/2. I was wondering which one is correct if any, and why? 回答1: In the simplest case a binary tree with a root node, a left and a right has 3 nodes, two of which are leaf nodes. It's (n+1)/2 . 回答2: If your total number nodes are n , and i are the total number of internal nodes ,i.e., whose degrees are 1. If the tree considered is a binary tree, then this relation holds true. 2i + 3 = n .