二叉树遍历(大部分)失败
Binary Tree Traversal (Mostly) Failing
对于class,我必须创建一个状态对象的二叉树,每个状态对象都包含一个常驻对象的二叉树,组织居住在每个状态的人。我正在尝试按名称在整个州树中搜索一个人(州树和居民树都是按名称的字母顺序排列的),这涉及遍历整个州树并在每个州的居民树中搜索该人。显然我的状态树遍历不起作用,因为大多数时候,当我知道一个事实时,它告诉我这个人不在数据库中(即我的 stateforperson 方法,如下所列,树 returns NULL)它们确实存在于数据库中。我确定我的 searchfor() 方法有效。
node <Person*> * stateforperson (string nm, node <T> * n)
{ if (n != NULL)
{
node <Person*> * person = n->data->residents->searchfor(nm);
if (person != NULL)
return person;
return stateforperson(nm, n->left);
return stateforperson(nm, n->right);
}
else
return NULL;
}
尝试更新:
node <Person*> * stateforperson (string nm, node <T> * n)
{ if (n != NULL)
{
node <Person*> * person = n->data->residents->searchfor(nm);
if (person != NULL)
return person;
// Here, you explore the left branch and get the results.
node <Person*> * left_ret = stateforperson(nm, n->left);
// Here, same with right branch.
node <Person*> * right_ret = stateforperson(nm, n->right);
// You now have both results.
if (left_ret != NULL) // If a result was found in left branches, you return that person.
return left_ret;
else if (right_ret != NULL) // Same with right branch.
return right_ret;
else // The problem was here. Before you returned uninitialized memory. (because there wasn't a specified return value.
// Now, you return a NULL pointer if nothing was found.
//So you detect that no person was found and don't use unitialized memeory.
return (NULL);
}
else
return NULL;
}
我相信发生的事情是你只探索了左边的节点。永远不要向右走,因为你 return 之前 :
return stateforperson(nm, n->left);
// Here you just return the left part. Never reaching the following line
return stateforperson(nm, n->right);
尝试实际存储值和更多类似的东西:
left_ret = stateforperson(nm, n->left);
right_ret = stateforperson(nm, n->right);
然后根据需要对变量进行任何检查以 return 正确。
(我相信这至少是问题所在。有一段时间没有进行任何递归编程,所以我可能会弄错。)
对于class,我必须创建一个状态对象的二叉树,每个状态对象都包含一个常驻对象的二叉树,组织居住在每个状态的人。我正在尝试按名称在整个州树中搜索一个人(州树和居民树都是按名称的字母顺序排列的),这涉及遍历整个州树并在每个州的居民树中搜索该人。显然我的状态树遍历不起作用,因为大多数时候,当我知道一个事实时,它告诉我这个人不在数据库中(即我的 stateforperson 方法,如下所列,树 returns NULL)它们确实存在于数据库中。我确定我的 searchfor() 方法有效。
node <Person*> * stateforperson (string nm, node <T> * n)
{ if (n != NULL)
{
node <Person*> * person = n->data->residents->searchfor(nm);
if (person != NULL)
return person;
return stateforperson(nm, n->left);
return stateforperson(nm, n->right);
}
else
return NULL;
}
尝试更新:
node <Person*> * stateforperson (string nm, node <T> * n)
{ if (n != NULL)
{
node <Person*> * person = n->data->residents->searchfor(nm);
if (person != NULL)
return person;
// Here, you explore the left branch and get the results.
node <Person*> * left_ret = stateforperson(nm, n->left);
// Here, same with right branch.
node <Person*> * right_ret = stateforperson(nm, n->right);
// You now have both results.
if (left_ret != NULL) // If a result was found in left branches, you return that person.
return left_ret;
else if (right_ret != NULL) // Same with right branch.
return right_ret;
else // The problem was here. Before you returned uninitialized memory. (because there wasn't a specified return value.
// Now, you return a NULL pointer if nothing was found.
//So you detect that no person was found and don't use unitialized memeory.
return (NULL);
}
else
return NULL;
}
我相信发生的事情是你只探索了左边的节点。永远不要向右走,因为你 return 之前 :
return stateforperson(nm, n->left);
// Here you just return the left part. Never reaching the following line
return stateforperson(nm, n->right);
尝试实际存储值和更多类似的东西:
left_ret = stateforperson(nm, n->left);
right_ret = stateforperson(nm, n->right);
然后根据需要对变量进行任何检查以 return 正确。
(我相信这至少是问题所在。有一段时间没有进行任何递归编程,所以我可能会弄错。)