tree-traversal

问题 I am writing my own Huffman encoder, and so far I have created the Huffman tree by using a minHeap to pop off the two lowest frequency nodes and make a node that links to them and then pushing the new node back one (lather, rinse, repeat until only one node). So now I have created the tree, but I need to use this tree to assign codes to each character. My problem is I don't know how to store the binary representation of a number in C++. I remember reading that unsigned char is the standard

问题 I came across solution given at http://discuss.joelonsoftware.com/default.asp?interview.11.780597.8 using Morris InOrder traversal using which we can find the median in O(n) time. But is it possible to achieve the same using O(logn) time? The same has been asked here - http://www.careercup.com/question?id=192816 回答1: If you also maintain the count of the number of left and right descendants of a node, you can do it in O(logN) time, by doing a search for the median position. In fact, you can

问题 Is there a way to do a DFT on a BeautifulSoup parse tree? I'm trying to do something like starting at the root, usually , get all the child elements and then for each child element get their children, etc until I hit a terminal node at which point I'll build my way back up the tree. Problem is I can't seem to find a method that will allow me to do this. I found the findChildren method but that seems to just put the entire page in a list multiple times with each subsequent entry getting

问题 How can I traverse an n -ary tree without using recursion? Recursive way: traverse(Node node) { if(node == null) return; for(Node child : node.getChilds()) { traverse(child); } } 回答1: You can do this without recursion and without a stack. But you need to add two extra pointers to the node: The parent node. So you can come back to the parent if you are finished. The current child node so you know which one to take next. For each node, you handle all the kids. If a kid is handled, you check if

问题 How can I traverse an n -ary tree without using recursion? Recursive way: traverse(Node node) { if(node == null) return; for(Node child : node.getChilds()) { traverse(child); } } 回答1: You can do this without recursion and without a stack. But you need to add two extra pointers to the node: The parent node. So you can come back to the parent if you are finished. The current child node so you know which one to take next. For each node, you handle all the kids. If a kid is handled, you check if

问题 I'm writing a simple function that walks through a directory tree looking for folders of a certain name. What I'm after is a match's parent path. E.g., for "C:/a/b/c/MATCH" I want "C:/a/b/c". I don't need duplicate parents or paths to subfolder matches, so if there's a "C:/a/b/c/d/e/f/MATCH", I don't need it. So, during my walk, once I have a parent, I want to iterate to the next current root. Below is what I have so far, including a comment where I'm stuck. def FindProjectSubfolders

问题 I'm writing a simple function that walks through a directory tree looking for folders of a certain name. What I'm after is a match's parent path. E.g., for "C:/a/b/c/MATCH" I want "C:/a/b/c". I don't need duplicate parents or paths to subfolder matches, so if there's a "C:/a/b/c/d/e/f/MATCH", I don't need it. So, during my walk, once I have a parent, I want to iterate to the next current root. Below is what I have so far, including a comment where I'm stuck. def FindProjectSubfolders

问题 I'm just starting with trees and am writing a function that traverses a binary tree and visits every node. I'm calling a function called doSomething(TreeNode *thisNode) for each node in the tree. I want to make sure if what I have is correct and that I'm on the right track? Thanks! void MyTree::Traverse(TreeNode *rt) { If(rt != NULL) Traverse(rt -> left); doSomething (rt); Traverse(rt -> right); } 回答1: Almost, but not quite. The if statement in C++ is not capitalized, and you must add

问题 Well, I know that a breadth-first-search-tree of an undirected graph can't have a back edge. But I'm wondering how can it even have a cross-edge? I'm not able to image a spanning tree of a graph G constructed out of OFS, that contains a cross-edge. 回答1: The process of building a spanning tree using BFS over an undirected graph would generate the following types of edges: Tree edges Cross edges (connecting vertices on different branches) A simple example: Imagine a triangle (a tri-vertice

问题 Ok, the question alone is making my head spin. I have an anchor tag that is calling a function: <a href="#" id="addPerson" onClick="addPerson(); return false;">Add a Guest</a> Just for reference, this is the function that's being called: function addPerson() { //current keeps track of how many rows we have. current++; console.log($(this).parent('form').attr('id')); var strToAdd = '<div class="row">\ <div class="column grid_2">\ <label for="two-guests-name'+current+'">Guest '+current+':</label