说我有在这个帖子，虽然它在我的情况下，一个载体，希望不应该的问题（他们在编程Clojure的书向量）：
Say I have a tree defined as per the recommendation in this post, although it's a vector in my case, which hopefully shouldn't matter (they're vectors in Programming Clojure book):
(def tree [1 [[2 [4] [5]] [3 [6]]]])
这应该是这样的：
which should be something like:
1 / \ 2 3 / \ | 4 5 6现在，我愿意做树的广度优先遍历没有任何的传统手段，如队列，而是使用专门的堆叠是传递信息。我知道这不是最简单的途径，但我主要是做锻炼。此外，在这一点上，我不打算返回一个集合（我会明白这一点之后的练习），但实际上只是打印出来的节点，因为我通过他们的旅行。
Now, I'd like to do a breadth-first traversal of the tree without any of the traditional means such as the queue, and instead use exclusively the stack to pass information around. I know this isn't the easiest route, but I'm doing it mostly as exercise. Also at this point I'm not planning to return a collection (I'll figure that out afterwards as exercise) but instead just print out the nodes as I travel through them.
我目前的解决方案（刚开始使用Clojure，是不错的）：
My current solution (just starting out with Clojure, be nice):
(defn breadth-recur [queue] (if (empty? queue) (println "Done!") (let [collections (first (filter coll? queue))] (do ; print out nodes on the current level, they will not be wrapped' ; in a [] vector and thus coll? will return false (doseq [node queue] (if (not (coll? node)) (println node))) (recur (reduce conj (first collections) (rest collections)))))))
最后一行并不如预期运行和我难倒关于如何解决它。我确切地知道我想要什么：我需要剥离载体的每一层，然后将结果连接到传递到易复发
The last line is not working as intended and I'm stumped about how to fix it. I know exactly what I want: I need to peel each layer of vectors and then concatenate the results to pass into recur.
我看到这个问题主要是一个：
The issue I'm seeing is mostly a:
IllegalArgumentException Don't know how to create ISeq from: java.lang.Long
基本上连接词不喜欢附加一个载体，一个长期的，如果我换连词对于 CONCAT ，然后我失败时，这两个项目我是一个串联的是不是一个向量。面对这两个连词和CONCAT失败：
Basically conj doesn't like appending a vector to a long, and if I swap conj for concat, then I fail when one of the two items I'm concatenating isn't a vector. Both conj and concat fail when facing:
[2 [4] [5] [3 [6]]]
我觉得我缺少一个真正的基本操作，在这里，将工作无论在载体和原语中的两个位置。
I feel like I'm missing a really basic operation here that would work both on vectors and primitives in both positions.
有什么建议？
修改1：
这棵树实际上应该是（Joost的感谢！）：
The tree should actually be (thanks Joost!):
(def tree [1 [2 [4] [5]] [3 [6]]])
不过，我们还没有找到一个宽度诚为先的解决方案。
However we still haven't found a breadth-first solution.
推荐答案由于显然还有很贴没有广度优先的解决方案，下面是一个简单的算法，来实现第一热切，然后转化为偷懒：
Since apparently there is still no breadth-first solution posted, here is a simple algorithm, implemented first eagerly, and then transformed to be lazy:
(defn bfs-eager [tree] (loop [ret [], queue (conj clojure.lang.PersistentQueue/EMPTY tree)] (if (seq queue) (let [[node & children] (peek queue)] (recur (conj ret node) (into (pop queue) children))) ret))) (defn bfs-lazy [tree] ((fn step [queue] (lazy-seq (when (seq queue) (let [[node & children] (peek queue)] (cons node (step (into (pop queue) children))))))) (conj clojure.lang.PersistentQueue/EMPTY tree)))更多推荐
难倒与功能广度优先遍历Clojure中？
发布评论