将二叉搜索树转化为排序的双向链表

链接：

​​​​​​LCR 155. 将二叉搜索树转化为排序的双向链表

题解：

/*
// 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) {
        _head = _tail = nullptr;
        dfs(root);
        if (_tail) {
            _tail->right = _head;
            _head->left = _tail;
        }
        return _head;
    }
private:
    void dfs(Node* root) {
        if (!root) {
            return;
        }
        dfs(root->left);
        if (_head == nullptr) {
            _head = root;
            _tail = root;
        } else {
            _tail->right = root;
            root->left = _tail;
            _tail = root;
        }
        dfs(root->right);
    }
private:
    Node* _head;
    Node* _tail;
};