树结构的深度优先遍历是应用中常见的问题
在实际项目中,多叉树出现的比较普遍,常用来存储类似字典词条的路径信息。
多叉树对于在一个序列中找到前缀匹配的所有路径是可行的选择,例如找到一段文字中所有前缀匹配的词条(中国人民解放军为例,有中,中国,中国人,中国人民,中国人民解放军等匹配词条)。
构造一棵包含所有中文词条的字典树,可以通过深度优先遍历快速解析出这些前缀匹配的词条,树的每一个节点都是一个汉字,尔从根节点出发的路径是存储的中文词条。
以下的代码是一段示例,它的遍历会输出所有的叶子节点。
树结构是一个名为Tree的类型模板,其中存储了TreeData类型的有效数据,使用定义的接口可以存储路径到Tree结构中。
遍历的类型TreeTraverser,使用了一个栈来记录遍历的当前路径,其类型是自定义的TraverStack类型,表示一个节点和它未被遍历的孩子,栈中每个节点未遍历的孩子的序列实现上使用了一个vector序列容器记录,便于回溯的时候访问下一个未访问的孩子节点。
#include<iostream> #include<stack> #include<queue> #include<cassert> #include<algorithm> using namespace std; template<class TreeData> struct Tree { private: static int TreeDataCompare(Tree* op1, Tree* op2) { return op1->GetData() - op2->GetData(); } public: Tree(TreeData data): m_data(data) { } public: void AppendChild(Tree* child) { m_children.push_back(child); // sort children trees by treedata sort(m_children.begin(), m_children.end(), TreeDataCompare); } Tree* PutChildByData(TreeData data) { // data already exists for(int i = 0; i < m_children.size(); i++) if(data == m_children[i]->GetData()) return m_children[i]; // Append new child to children array Tree* newChild = new Tree(data); AppendChild(newChild); return newChild; } void PutPathByDataArray(const TreeData* szData) { if (*szData == 0) return; Tree* child = PutChildByData(*szData); child->PutPathByDataArray(szData+1); } private: TreeData m_data; vector<Tree*> m_children; public: int GetChildrenNum() { return m_children.size(); } Tree* GetChildByIndex(int index) { return m_children[index]; } TreeData GetData() { return m_data; } // Fill children to the specified queue virtual void FillQueueWithChildren(queue<Tree*>& queue) { for(int i = 0; i < m_children.size(); i++) { if(m_children[i]) queue.push(m_children[i]); } } }; template<class Tree> class TraverseStack { public: TraverseStack(Tree* tree): m_tree(tree) { m_tree->FillQueueWithChildren(m_children); } Tree* GetNextChild() { if (m_children.empty()) return NULL; // pop head of the untraversed children queue Tree* head = m_children.front(); m_children.pop(); return head; } Tree* GetTree() { return m_tree; } private: Tree* m_tree; queue<Tree*> m_children; }; template<class Tree> class BFSTraverser { public: BFSTraverser(Tree* root):m_root(root){} virtual ~BFSTraverser(){} public: typedef TraverseStack<Tree> PATHSTACKITEM; typedef vector<PATHSTACKITEM > PATHSTACK; public: virtual void Traverse() { m_pathStack.clear(); // push the root stack item PATHSTACKITEM rItem(m_root); m_pathStack.push_back(rItem); while(!m_pathStack.empty()) { PATHSTACKITEM& top = m_pathStack.back(); //cout << "Get top = " << top.GetTree()->GetData() << endl; Tree* nextChild = top.GetNextChild(); if (!nextChild) { // output pathStack if(top.GetTree()->GetChildrenNum() == 0) OutputStack(); // go back along the path to parent TraverseStack element m_pathStack.pop_back(); continue; } assert(nextChild); // pre order output root‘s path if(nextChild == top.GetTree()->GetChildByIndex(0)) ;//OutputStack(); // push new TraverseStack element to path PATHSTACKITEM newStackItem(nextChild); // enlonger the current path to the untraversed child m_pathStack.push_back(newStackItem); continue; } } private: void OutputStack() { for(int i = 1; i < m_pathStack.size(); i++) { if(i>0) ;//cout << ","; cout << m_pathStack[i].GetTree()->GetData(); } cout << endl; } private: Tree* m_root; PATHSTACK m_pathStack; };
时间: 2024-12-27 21:45:02