在BST中实现向后迭代器

给定二进制搜索树，任务是使用以下功能在其上实现向后迭代器。

1. curr()：返回指向当前元素的指针。
2. prev()：迭代到二进制搜索树中的前一个最大元素。
3. isEnd()：如果没有要遍历的节点，则返回true，否则返回false。

迭代器以降序遍历BST。我们将使用堆栈数据结构实现迭代器。
初始化：

• We will create a stack named “q” to store the nodes of BST.
• Create a variable “curr” and initialise it with pointer to root.
• While “curr” is not NULL 
  • Push “curr” in the stack ‘q’.
  • Set curr = curr -> right

curr()

Returns the value at the top of the stack ‘q’. 
Note: It might throw segmentation fault if the stack is empty. 
 

时间复杂度： O(1)
prev()

• Declare pointer variable “curr” which points to node.
• Set curr = q.top()->left.
• Pop top most element of stack.
• While “curr” is not NULL 
  • Push “curr” in the stack ‘q’.
  • Set curr = curr -> right.

时间复杂度：平均所有通话为O(1)。在最坏的情况下，单次通话可以为O(h)。
isEnd()

Returns true if stack “q” is empty else returns false. 
 

时间复杂度： O(1)
此迭代器实现的最坏情况下，空间复杂度为O(h)。应该注意的是
迭代器指向堆栈的最顶层元素。
下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
 
// Node of the binary tree
struct node {
    int data;
    node* left;
    node* right;
    node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};
 
// Iterator for BST
class bstit {
private:
    // Stack to store the nodes
    // of BST
    stack q;
 
public:
    // Constructor for the class
    bstit(node* root)
    {
        // Initializing stack
        node* curr = root;
        while (curr != NULL)
            q.push(curr), curr = curr->right;
    }
 
    // Function to return
    // current element iterator
    // is pointing to
    node* curr()
    {
        return q.top();
    }
 
    // Function to iterate to previous
    // element of BST
    void prev()
    {
        node* curr = q.top()->left;
        q.pop();
        while (curr != NULL)
            q.push(curr), curr = curr->right;
    }
 
    // Function to check if
    // stack is empty
    bool isEnd()
    {
        return !(q.size());
    }
};
 
// Function to test the iterator
void iterate(bstit it)
{
    while (!it.isEnd())
        cout << it.curr()->data << " ", it.prev();
}
 
// Driver code
int main()
{
    node* root = new node(5);
    root->left = new node(3);
    root->right = new node(7);
    root->left->left = new node(2);
    root->left->right = new node(4);
    root->right->left = new node(6);
    root->right->right = new node(8);
 
    // Iterator to BST
    bstit it(root);
 
    // Function call to test the iterator
    iterate(it);
 
    return 0;
}

输出：

8 7 6 5 4 3 2

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