力扣链接https://leetcode.cn/problems/er-cha-sou-suo-shu-yu-shuang-xiang-lian-biao-lcof/description/

解题思路

中序遍历可以得到排序的序列，遍历时维护当前节点和前一个节点，改变左右指针。头节点设为null，遍历开始判断一下然后初始化。尾节点中序遍历时一直更新。

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val,Node _left,Node _right) {
        val = _val;
        left = _left;
        right = _right;
    }
};
*/

class Solution {
    Node head,end;
    void dfs(Node root){
        if(root==null){
            return;
        }
        dfs(root.left);
        if(head==null){
            head=root;
        }
        else{
            root.left=end;
            end.right=root;
        }
        end=root;        //作为前继节点
        dfs(root.right);
    }
    public Node treeToDoublyList(Node root) {
        if(root==null){
            return null;
        }
        dfs(root);
        head.left=end;
        end.right=head;
        return head;
    }
}