该类的目的是模拟二叉搜索树的功能。在下面的代码中,我试图将它从一个struct和一堆函数改编成一个名为BST的包装类。然而,我不确定的一件事是如何从节点结构中访问“root”。 Root目前在BST类中声明。
class bst
{
public:
struct Node
{
public:
int data;
struct Node *left;
struct Node *right;
Node* FindMin(Node* root)
{
while(root->left != NULL) root = root->left;
return root;
}
Node* Insert(Node *root,int data)
{
if(root == NULL) {
root = new Node();
root->data = data;
root->left = root->right = NULL;
//Update Height & Size
bstHeight = 0;
bstSize = 0;
}
else if(data <= root->data)
root->left = Insert(root->left,data);
else
root->right = Insert(root->right,data);
return root;
}
Node* Delete(struct Node *root, int data)
{
if(root == NULL) return root;
else if(data < root->data) root->left = Delete(root->left,data);
else if (data > root->data) root->right = Delete(root->right,data);
//Value found
else {
// Case 1: No child
if(root->left == NULL && root->right == NULL)
{
delete root;
root = NULL;
//Update Height & Size
bstHeight = 0;
bstSize = 0;
}
//Case 2: One child
else if(root->left == NULL)
{
struct Node *temp = root;
root = root->right;
delete temp;
//Update Height & Size
bstHeight = 0;
bstSize = 0;
}
else if(root->right == NULL)
{
struct Node *temp = root;
root = root->left;
delete temp;
//Update Height & Size
bstHeight = 0;
bstSize = 0;
}
// case 3: 2 children
else
{
struct Node *temp = FindMin(root->right);
root->data = temp->data;
root->right = Delete(root->right,temp->data);
//Update Height & Size
bstHeight = 0;
bstSize = 0;
}
}
return root;
}
//# of Nodes in tree
void size(Node *root)
{
//Check if end
if(root == NULL) return;
//Not end
else
{
bstSize = bstSize + 1;
size(root->left); //Visit left subtree
size(root->right); // Visit right subtree
}
}
void height(Node *root, int temp)
{
//Check if end
if(root == NULL)
{
if(temp > bstHeight)
{
bstHeight = temp;
}
return;
}
//Not end
else
{
temp = temp + 1;
height(root->left, temp); //Visit left subtree
height(root->right, temp); // Visit right subtree
}
}
//Function to visit nodes in Inorder
void show()
{
if(root == NULL) return;
show(root->left); //Visit left subtree
printf("%d ",root->data); //Print data
show(root->right); // Visit right subtree
}
void check(Node *root)
{
//End of a 'branch'
if(root == NULL) return;
int value = 0;
value = root->data;
//Checking left subtree
if(value < root->left->data)
{
//Tree is NOT valid
valid = 0;
}
//Checking right subtree
if(value > root->right->data)
{
//Tree is NOT valid
valid = 0;
}
check(root->left); //Visit left subtree
printf("%d ",root->data); //Print data
//check(root->right); // Visit right subtree
}
};
Node* root = NULL;
};
具体来说,在show函数中。它并不像使用其他函数将其放入Node那么简单,因为root需要是唯一的,并且新Node至少被调用一次。 Show不会在当前状态下编译,我不知道从哪里开始。
答案 0 :(得分:0)
虽然评论说明了所有内容,但我还要提一下:
当你想让你的代码尽可能地与你现有的代码保持相似时,尝试向Node类添加一个构造函数,该构造函数需要一个指针或引用(最好)到root并检查,每次你创建一个节点,您将根提供给构造函数。
顺便说一下,在C ++中查看一些简单的基于Node的数据结构实现甚至可能是更好的方法,例如在线程中