我正在尝试在键中查找名称。我认为它恢复得很好。但是,它出现时未找到。也许我的代码某处有误?
if (database.retrieve(name, aData)) // both contain the match
在 main()
中
static void retrieveItem(char *name, data& aData)
{
cout << ">>> retrieve " << name << endl << endl;
if (database.retrieve(name, aData)) // name and aData both contain the match
cout << aData << endl;
else
cout << "not found\n";
cout << endl;
}
static void removeItem(char *name)
{
cout << ">>> remove " << name << endl << endl;
if (database.remove(name))
cout << name << " removed\n";
else
cout << name << " not found\n";
cout << endl;
}
int main()
{
#ifdef _WIN32
// request memory leak report in Output Window after main returns
_CrtSetDbgFlag ( _CRTDBG_ALLOC_MEM_DF | _CRTDBG_LEAK_CHECK_DF );
#endif
data aData;
<< "Database Of Great Computer Scientists\n\n";
database.insert(data("Ralston, Anthony"));
database.insert(data("Liang, Li"));
database.insert(data("Jones, Doug"));
database.insert(data("Goble, Colin"));
database.insert(data("Knuth, Donald"));
database.insert(data("Kay, Alan"));
database.insert(data("Von Neumann, John"));
database.insert(data("Trigoboff, Michael"));
database.insert(data("Turing, Alan"));
displayDatabase(true);
retrieveItem("Trigoboff, Michael", aData);
retrieveItem("Kaye, Danny", aData);
removeItem("Ralston, Anthony");
displayDatabase(true);
检索函数...
bool BST::retrieve(const char *key, data &aData, int parent) const
{
for(int index=0; index < maxsize+1; index++)
{
if (!items[index].empty)
{
if ( items[index].instanceData == key )
{
aData.setName(key);
return true; // doesn't return right away
}
}
}
}
并在data.cpp中定义
bool operator== (const data& d1, const data& d2)
{
return strcmp(d1.getName(), d2.getName()) == 0;
}
所以当我认为它应该正常工作时,main() 中的这段代码就是它说找不到的地方。 name 和 aData 都包含找到的正确名称..
static void retrieveItem(char *name, data& aData)
{
cout << ">>> retrieve " << name << endl << endl;
if (database.retrieve(name, aData)) // name and aData both contain the match
cout << aData << endl;
else
cout << "not found\n";
cout << endl;
}
最佳答案
您应该使用 BST 在树中导航 - 而不是像其他人所说的那样遍历数组中的每个项目。尝试类似的东西:
bool retrieve(key, aData)
retrieve(key, aData, parent)
if (key == aData)
return true
else
return false
bool retrieve(key, aData, parent)
if (key == items[parent].name)
aData.setName(key)
else if (key < items[parent].name)
retrieve(key, aData, 2*parent+1)
else
retrieve(key, aData, 2*parent+2)
那应该很好用! :)
关于c++ - 搜索二叉搜索树,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1853082/