经过几个问题和一些很好的答案和友好的帮手在这里。我在二元树中删除了我的问题，我得到了建议，我不能只删除树中最大的数字，因为它可能不是最后一个，或者它有儿童1,2或没有，所以我做了下面的代码，我用了很多评论希望，可以帮助你的人帮助我。我现在真的不知道的是，我如何在我的公共中调用此RemoveLargest()函数，然后在main中调用，即使我不知道代码是否能正常运行。
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
template<class T>
class BinaryTree
{
struct Node
{
T data;
Node* lChildptr;
Node* rChildptr;
Node(T dataNew)
{
data = dataNew;
lChildptr = NULL;
rChildptr = NULL;
}
};
private:
Node* root;
void Insert(T newData, Node* &theRoot) //Insert elements into the tree start.
{
if(theRoot == NULL)
{
theRoot = new Node(newData);
return;
}
if(newData < theRoot->data)
Insert(newData, theRoot->lChildptr);
else
Insert(newData, theRoot->rChildptr);
} //end.
void PrintTree(Node* theRoot) //print me the tree /start
{
if(theRoot != NULL)
{
PrintTree(theRoot->lChildptr);
cout<< theRoot->data<<" \n";
PrintTree(theRoot->rChildptr);
}
} //end.
T Largest( Node* theRoot) // show me largest number /start.
{
if ( root == NULL )
{
cout<<"There is no tree";
return -1;
}
if (theRoot->rChildptr != NULL)
{
return Largest(theRoot->rChildptr);
}
T value = theRoot->data;
return value;
} //end.
void RemoveLargest(Node* theRoot) //remove the largest priority number from tree /start.
{
Node* current; //the current tree?
Node* parent; //the parent of the current node?
current=theRoot;
// 3 cases :
// 1. We're removing a leaf node
// 2. We're removing a node with a single child
// 3. we're removing a node with 2 children
//Node with single child.
if((current->lChildptr == NULL && current->rChildptr != NULL)||(current->lChildptr != NULL && current->rChildptr == NULL))
{
if(current->lChildptr == NULL && current->rChildptr != NULL)
{
if(parent->lChildptr==current)
{
parent->lChildptr = current->rChildptr;
delete current;
}
else
{
parent->rChildptr = current->rChildptr;
delete current;
}
}
else //left child ok, no right child
{
if(parent->lChildptr==current)
{
parent->lChildptr = current->lChildptr;
delete current;
}
else
{
parent->rChildptr = current->lChildptr;
delete current;
}
}
return;
}
//We found a leaf(a node with not a single child)
if(current->lChildptr == NULL && current->rChildptr == NULL)
{
if (parent->lChildptr == current)
parent->lChildptr = NULL;
else
parent->rChildptr = NULL;
delete current;
return;
}
//Node with 2 children
// replace node with smallest value in right subtree
if (current->lChildptr != NULL && current->rChildptr != NULL)
{
Node* checkr;
checkr = current->rChildptr;
if((checkr->lChildptr == NULL)&&(checkr->rChildptr == NULL))
{
current=checkr;
delete checkr;
current->rChildptr = NULL;
}
else //right child has children
{
//if the node's right child has a left child
//Move all the way down left to locate smallest element
if ((current->rChildptr)->lChildptr != NULL)
{
Node* lcurr;
Node* lcurrp;
lcurrp = current->rChildptr;
lcurr = (current->rChildptr)->lChildptr;
while(lcurr->lChildptr != NULL)
{
lcurrp = lcurr;
lcurr = lcurr->lChildptr;
}
current->data = lcurr->data;
delete lcurr;
lcurrp->lChildptr = NULL;
}
else
{
Node* temp;
temp = current->rChildptr;
current->data = temp ->data;
current->rChildptr = temp->rChildptr;
delete temp;
}
}
return;
}
};
public:
BinaryTree()
{
root = NULL;
}
void AddItem(T newData)
{
Insert(newData, root);
}
void PrintTree()
{
PrintTree(root);
}
T Largest()
{
return Largest(root);
}
void RemoveLargest()
{
RemoveLargest();
}
};
int main()
{
BinaryTree<int> *myBT = new BinaryTree<int>();
myBT->AddItem(5);
myBT->AddItem(1);
myBT->AddItem(4);
myBT->AddItem(2);
myBT->AddItem(3);
//for(int i = 0; i < 10; i++) //randommal tolti fel/fill with random
//myBT->AddItem(rand() % 100);
cout << "BinaryTree:" << endl; //kilistazaa a fat/ list my tree
myBT->PrintTree();
cout << "Largest element: " << myBT->Largest() << endl; //visszaadja a legnagyobb elemet/shows the largest number
myBT->RemoveLargest(); //suposed to delet the largest number
myBT->PrintTree(); //shows list again
}
编辑了代码，现在它正在运行，它创建的树，显示最大，但在我调用后我的删除功能仍然崩溃... = /
答案 0 :(得分：1)
就像我之前说过的那样，我认为你的事情过于复杂。您必须考虑在二进制搜索树的上下文中节点是最大的节点意味着什么，以及节点中的键之间的关系。
如果节点是树中最大的节点，则它不可能具有正确的子指针，因为正确的子节点必须具有更大的键。然后，如果你知道它最多只有一个左子，你只需用你可能为空的左子项代替你的节点，你就完成了。
T ExtractLargest(Node*& n){
if (!n)
return -1;
if (n->rChildptr)
return ExtractLargest(n->rChildptr);
T result = n->data;
Node *d = n;
n = n->lChildptr;
delete d;
return result;
}