Question

我有一张以下形式的树：

Node[0]:
       type: "element",
       name: "div",
       data: "",
       attributes:{"class":"wrapper"},
       children: 
                Node[0]:
                        type: "text",
                        name: "",
                        data: "Hello",
                        attributes: {},
                        children: null,
                Node[1]:
                        type: "element",
                        name: "span",
                        data: "",
                        attributes: {"class:leftMenu", "class:Applyshadow"},
                        children: 
                                 Node[0]:
                                        type: "text",
                                        name: "",
                                        data: "World!",
                                        attributes: {},
                                        children: null
Node[1]:
       type: "element",
       name: "div",
       data: "",
       attributes: {"class":"secondDiv", "id":"submit"},
       children: null

“element”类型的节点可以有子节点，如果是，则将子节点存储为节点向量。我正在尝试使用以下类重建树：

struct Attribute{
    std::string name;
    std::string value;
};

struct Node{
    std::string type;
    std::string name;
    std::string data;
    std::vector<Attribute> attributes;
    std::vector<Node> children;
 };


class HtmlTree{
    public:
        std::vector<Node> nodes;
        void buildTree(GumboNode*)
}

树结构来自gumboNode，它是gumboParser的一部分，在buildTree方法的实现中，我能够打印树中的所有节点，但仍然坚持如何将它存储在nodes向量中。 / p>

Void HtmlTree::buildTree(GumboNode* root){
    print root->type
    vector children = root->children;
    for each child in children:
        if child->type == "element":
            print child->type;
            buildTree(child)
        elif child->type == "text":
            print child->type;
}

上面的Pseudocode打印树中所有节点的类型。有人可以帮助我利用相同的递归方法来存储向量nodes中的所有节点吗？

Answer 1

你有这样的想法吗？

void HtmlTree::buildTree(GumboNode *root) {
    nodes = buildTreeImp(root) ;
}

Node HtmlTree::buildTreeImp(GumboNode* root){
    Node result ;
    vector children = root->children;
    for each child in children:
        if child->type == "element": {
            nodes.push_back(buildTree(child))
        elif child->type == "text": {
            Node text_node ;
            text_node.type = child.type ;   
            // set other attributes
            nodes.push_back(text_node) ;
    } 
}

重建一棵树

1 个答案: