trie这种树也被称为线索，搜索树。

正如图

以下是用stl 的map来实现

class trie_item_c
{
public:trie_item_c(){}trie_item_c(const char nm){name = nm;}void set_name(const char nm){name = nm;}trie_item_c * get_child(const char nm){map<const char ,trie_item_c*>::const_iterator it = children.find(nm);if(it != children.end())return it->second;else{trie_item_c *cld = new trie_item_c(nm);children.insert(pair <const char ,trie_item_c*>(nm,cld));return cld;}}bool find(const char* dic){if(!dic || *dic == '\0'){if(children.end() != children.find('\0'))return true;return false;}const char* temp = dic;map<const char ,trie_item_c*>::const_iterator it = children.find(*temp);if(it == children.end())return false;return it->second->find(++temp);}void print(){printf("%c",name);map<const char ,trie_item_c*>::const_iterator it = children.begin();for(;it != children.end();it++){it->second->print();}printf("\n");}virtual ~trie_item_c(){map<const char ,trie_item_c*>::const_iterator it = children.begin();while(it != children.end()){delete it->second;children.erase(it);it = children.begin();}}
private:map<const char ,trie_item_c*> children;char name;};

以下代码是构建树的过程

for(const char* dic = lst.first_string();dic;dic = lst.next_string()){trie_item_c *temp = root;for(int i=0;i<str_temp.str_len();i++){char c = str_temp.get(i);trie_item_c *item = temp->get_child(c);temp = item;}//add one null childtemp->get_child('\0');}

推断是否一个字root在，只需要调用root->find("book");您可以。