樹形結構的路徑樹

Question

你好，假設你有普通的unix路徑樹作爲輸入（如字符串）。

root 0 
root/file1.txt 1 
root/file2.txt 2 
root/folder1 3 
root/folder1/file3.txt 4 
root/folder1/file4.txt 5 
e.t.c. 

將此字符串轉換爲樹型數據結構的更好方法是什麼？

Answer 1

現在我用簡單的樹表示，通過創建自己根節點

template<typename T> 
class TreeNode 
{ 
public: 
    TreeNode() {}; 

    TreeNode(T) 
    { 
     value = T; 
    } 

    TreeNode(const T& value) 
     : Value(value) 
    { 
    } 

    T Value; 
    vector<TreeNode<T>*> Children; 
}; 

我真的不明白吉爾的算法（上面貼出）。但我想解決方案必須如下：

1. Get a root node, set depth_level = 0 
3. set support_node = root_node 
4. for each path line 
5.  determine the quantity of slashes "/", key and file(folder) name 

so for example in string root/folder1/file4.txt 5, num of slashes = 2 filename = file4.txt, key = 5 
     create current_node 
6.  if num_of_slashes == level + 2 
7.   set support_node = current_node 
8.  if num_of_slashes == level + 1 
9.   add children to support_node 
10. And after that we must remember all ancestors, going down to leaves. Cuz we can return to any ancestor. 

對我來說這個算法似乎是非常複雜的。我不明白算法，上面發佈，也許有可能澄清這個問題？也許我用來存儲樹的結構不是最好的？

Answer 2

是這樣的：

創造 '\'

node=root; 
token=getNextToken(inputString); 
while (token){ 
if (!node.childExist(token)) node.createChild(token) 
node=node.Child(token) 
token=getNextToken(inputString); 
}