[英]Unordered Binary Tree Traversal
For class, I have to create a binary tree of state objects, each of which includes a binary tree of resident objects organizing the people who live in each state. 对于类,我必须创建一个状态对象的二叉树,每个对象都包括一个常驻对象的二叉树,这些居民对象组织着生活在每个州的人们。 I'm trying to search a given state for its oldest resident;
我正在尝试搜索给定州的最老居民; however, the residents are organized in the tree by alphabetical order, which does absolutely nothing for my search.
但是,居民是按字母顺序排列在树上的,这对我的搜索绝对没有任何帮助。 Thus, I have to traverse the entire tree of residents, updating the node that saves the oldest person, and return it once the tree has been fully traversed.
因此,我必须遍历居民的整个树,更新保存最老者的节点,并在遍历所有树后将其返回。 I have the first part of my code but am stuck on how to write the rest of the recursion.
我有我的代码的第一部分,但仍然停留在如何编写其余的递归上。
The state tree's method: 状态树的方法:
node <Person*> * findoldest (int obd, node <Person*> * oldest, node <Person*> * n)
{
//FINAL WORKING CODE
if (root == NULL)
return NULL;
if (n == NULL)
return NULL;
else
{
if (n->data->birthday < obd)
{
obd = n->data->birthday;
oldest = n;
}
node <Person*> * o_left = findoldest(obd, oldest, n->left);
node <Person*> * o_right = findoldest(obd, oldest, n->right);
node <Person*> * res;
if (o_right && o_left)
if (o_right->data->birthday < o_left->data->birthday)
res = o_right;
else
res = o_left;
else
res = (o_right != NULL ? o_right : o_left);
if (res && oldest)
if (res->data->birthday < oldest->data->birthday)
return res;
else
return oldest;
else
return ((res != NULL ? res : oldest));
}
}
And then the public "wrapper" state tree method: 然后是公共的“包装器”状态树方法:
node <Person*> * findoldest ()
{ int oldest_bday = root->data->birthday;
node <Person*> * oldest_person = root;
findoldest(oldest_bday, oldest_person, root);
}
This the pseudo-code you need: 这是您需要的伪代码:
node <Person*> * findoldest (node <Person*> * n)
{
if n->right != null :
right_oldest = findoldest(n->right)
if n->left != null:
left_oldest = findoldest(n->left)
return the node that has max value in (right_oldest.data.birthday, left_oldest.data.birthday, n.data.birthday)
}
Essentially, it's the same answer as your last post. 本质上,它与您上一篇文章的答案相同。
right_old = findoldestn(n->right);
left_old = findoldestn(n->left);
then figure out the oldest one between left/right and current, and return that value. 然后找出左/右和当前之间最早的一个,并返回该值。 And that can be put in place with
这可以放在适当的位置
res = (right_old->age > left_old->age ? right_old : left_old);
finalRet = (res->age > oldest->age ? res : oldest);
return (finalRet);
Or an equivalent with if notation : 或等效的if符号:
if (right_old->age >left_old->age)
res = right_old;
else
res = left_old;
if (res->age > oldest->age)
finalRes = res;
else
finalRes = oldest;
Fyi, i'm lazy, variable->age is equivalent to variable->data->birthday. Fyi,我很懒,variable-> age等同于variable-> data-> birthday。
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