426. Convert Binary Search Tree to Sorted Doubly Linked List

inorder dfs O(n) time O(h) space
维护一个prev用root更新prev，因为prev一直被更新，所以到最后prev就是tail，最后连接head和prev（即tail）即可

/*
// 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; // 注意这里必须提前判空，因为后边不好处理
        dfs(root);
        auto head = dummy_head.right;
        head->left = prev;
        prev->right = head;
        return head;
    }

    void dfs(Node *root) {
        if (!root) return;
        dfs(root->left);
        prev->right = root;
        root->left = prev;
        prev = root;
        dfs(root->right);
    }

    Node dummy_head, *prev = &dummy_head;
};