426. Convert Binary Search Tree to Sorted Doubly Linked List

问题描述：

Convert a BST to a sorted circular doubly-linked list in-place. Think of the left and right pointers as synonymous to the previous and next pointers in a doubly-linked list.

Let's take the following BST as an example, it may help you understand the problem better:

We want to transform this BST into a circular doubly linked list. Each node in a doubly linked list has a predecessor and successor. For a circular doubly linked list, the predecessor of the first element is the last element, and the successor of the last element is the first element.

The figure below shows the circular doubly linked list for the BST above. The "head" symbol means the node it points to is the smallest element of the linked list.

Specifically, we want to do the transformation in place. After the transformation, the left pointer of the tree node should point to its predecessor, and the right pointer should point to its successor. We should return the pointer to the first element of the linked list.

The figure below shows the transformed BST. The solid line indicates the successor relationship, while the dashed line means the predecessor relationship.

解题思路：

这道题目实际上也是将二叉搜索树序列化。

注意转换规则：左指针指向前继节点，右指针指向后继节点。

实际上就是中序遍历。

使用栈来帮助我们进行中序遍历。

在出栈时改变它的左指针，指向，前继节点prev；prev的右指针指向当前出栈的节点。

需要注意的是：

　　在最后需要连接头指针和尾指针。

代码：

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;

    Node() {}

    Node(int _val, Node* _left, Node* _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
public:
    Node* treeToDoublyList(Node* root) {
        if(!root) return root;
        Node* cur = root;
        stack<Node*> stk;
        Node* head = NULL;
        Node* prev = NULL;
        while(cur || !stk.empty()){
            if(cur){
                stk.push(cur);
                cur = cur->left;
            }else{
                cur = stk.top();
                stk.pop();
                if(!head) head = cur;
                if(prev){
                    prev->right = cur;
                    cur->left = prev;
                }
                prev = cur;
                cur = cur->right;
            }
        }
        head->left = prev;
        prev->right = head;
        return head;
    }
};