力扣-剑指Offer 36题 二叉搜索树与双向链表（C++）- dfs

题目链接：https://leetcode-cn.com/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/
题目如下：

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

    Node() {}

    Node(int _val) {
        val = _val;
        left = NULL;
        right = NULL;
    }

    Node(int _val, Node* _left, Node* _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/
class Solution {
public:
    Node* treeToDoublyList(Node* root) {
        if(!root) return root;
        //二叉搜索树中序是有序的
        dfs(root);//
        //尾指向头，头指向尾 顺时针方式指
        tail->right=dummy->right;
        dummy->right->left=tail;
        return dummy->right;
    }
    void dfs(Node* root){
        if(!root) return;

        Node* left=root->left;
        Node* right=root->right;

        dfs(root->left);
        
        //在循环双链表中尾插节点 顺时针方式指
        tail->right=root;
        root->left=tail;
        tail=root;

        dfs(root->right);
    }
private:
    Node* dummy=new Node();
    Node* tail=dummy;//需要尾插节点
};