traversal

问题 Say I have the path gui/site/junior/profile.py How do I get this?: gui gui/site gui/site/junior Bonus if you tell me how to loop through each path :D 回答1: You can loop with awk: awk 'BEGIN{FS=OFS="/"} { for (i=1; i<=NF; i++) { for (j=1; j<i; j++) printf "%s/", $j printf "%s\n", $i } }' <<< "gui/site/junior/profile.py" See as one liner: $ awk 'BEGIN{FS=OFS="/"}{for (i=1; i<=NF; i++) { for (j=1; j<i; j++) printf "%s%s", $j, OFS; printf "%s\n", $i}}' <<< "gui/site/junior/profile.py" gui gui/site

问题 I am having issues with selecting a certain element in my html When I click on the link with class "event_rsvp", I want to effect the HTML of the span in the next li with the class of "interested-status" I have tried closest, tried going out to the parents, how do I tell one to talk to the other. <li rel="101590"><span class="rsvp-status">Are You Going? – <a href="javascript:;" class="event_rsvp" rel="attending">RSVP</a></span></li> <li rel="101590"><span class="interested-status"> <a href=

问题 I have a vector based binary tree and need to apply a function to each value in the tree using various methods of traversal. The preorder traversal was very easy to implement with a recursive function but I have been having trouble doing the same with the inorder and postorder traversals. If anyone could help out that would be great! Some extra information that I should have included: I am using a vector of nodes, each node containing a boolean variable stating whether or not that node is

问题 I've been scratching my head for several hours for this... problem: Binary Tree (0) depth 0 / \ 10 20 depth 1 / \ / \ 30 40 50 60 depth 2 I am trying to write a function that takes depth as argument and return the sum of values of nodes of the given depth. For instance, if I pass 2, it should return 180 (i.e. 30+40+50+60) I decided to use breath first search and when I find the node with desired depth, sum up the value, but I just can't figure out how to find out the way which node is in what

问题 I want to calculate the security cost of middlebox traversal when VM migrate from one physical server to another. Middle boxes can be firewalls or IPS/IDS containing rules checking the VM traversing them. Now imagine the most simple scenario that the only problem is to find the cost of checking VM by middlebox rules (this is what I call it security cost), and according to this cost finding the optimum path. However there are already some protocols out there such as BGP or OSPF, but

问题 This question already has an answer here : Closed 7 years ago . Possible Duplicate: PHP - Iterate through folders and display HTML contents Using PHP, I'm trying to create a script that will navigate to the root directory of a website, and from there, using scandir() or glob() (those being the only directory scanning functions I've learned), I would use a recursive method to navigate through all the items in the root directory, then re-calling itself when encountering an entry that tested to

问题 While I traverse a nested Dict in Julia, it gives this Error: ERROR: access to undefined reference in next at dict.jl:567 Here is the code where you can reproduce this error: a = [0,19620,7291,32633,9,32513,42455,10045,31964,42455,11767,54] b = [14318,16405,19,18913,19,8141,18958,12336,7,16588,17358,30] d = Dict() for aa in a for bb in b if ! haskey(d,aa) d[aa]=Dict() end d[aa][bb] = 0.5 end end for k1 in keys(d) s =0.0 for k2 in keys(d[k1]) s+= d[k1][k2] end for k2 in keys(d[k1]) d[k1][k2] =

问题 I have a simple tree, defined thusly: type BspTree = | Node of Rect * BspTree * BspTree | Null I can get a collection of leaf nodes (rooms in my tree "dungeon") like this, and it seems to work: let isRoom = function | Node(_, Null, Null) -> true | _ -> false let rec getRooms dungeon = if isRoom dungeon then seq { yield dungeon } else match dungeon with | Node(_, left, right) -> seq { for r in getRooms left -> r for r in getRooms right -> r } | Null -> Seq.empty But now I'm trying to get

问题 I have a simple tree, defined thusly: type BspTree = | Node of Rect * BspTree * BspTree | Null I can get a collection of leaf nodes (rooms in my tree "dungeon") like this, and it seems to work: let isRoom = function | Node(_, Null, Null) -> true | _ -> false let rec getRooms dungeon = if isRoom dungeon then seq { yield dungeon } else match dungeon with | Node(_, left, right) -> seq { for r in getRooms left -> r for r in getRooms right -> r } | Null -> Seq.empty But now I'm trying to get

问题 I was trying to understand how it is so intuitive to implement postorder traversal using 2 stacks. How did someone come up with it, is it just an observation or some particular way of thinking which helps one come up with such methods. If yes then please explain how to think in the right direction. 回答1: Let me explain how I stumbled on the solution: You start off with this simple observation: prima facie, the pre-order and post-order traversal differ only in their order of operations :