這不是一個(gè)Leetcode 題目.我自己寫了下。
題意如下：
Convert a binary search tree to a sorted, circular, doubly-linked list, in place (using the tree nodes as the new list nodes).

use leftChild as "prev"

use rightChild as "next

My code:

import java.util.Comparator;
import java.util.PriorityQueue;

public class Solution {
    public TreeNode traverse(TreeNode root) {
        if (root == null)
            return null;
        return dfs(root)[0];
    }
    
    private TreeNode[] dfs(TreeNode root) {
        if (root == null)
            return null;
        TreeNode[] left = dfs(root.left);
        TreeNode head = null;
        TreeNode tail = null;
        if (left != null) {
            head = left[0];
            left[1].right = root;
            root.left = left[1];
        }
        TreeNode[] right = dfs(root.right);
        if (right != null) {
            root.right = right[0];
            right[0].left = root;
            tail = right[1];
        }
        TreeNode[] ret = new TreeNode[2];
        ret[0] = (head == null ? root : head);
        ret[1] = (tail == null ? root : tail);
        return ret;
    }
    
    
    
    public static void main(String[] args) {
        Solution test = new Solution();
        TreeNode a1 = new TreeNode(1);
        TreeNode a2 = new TreeNode(2);
        TreeNode a3 = new TreeNode(3);
        TreeNode a4 = new TreeNode(4);
        TreeNode a5 = new TreeNode(5);
        TreeNode a6 = new TreeNode(6);
        TreeNode a7 = new TreeNode(7);
        a1.left = a2;
        a1.right = a3;
        a2.left = a4;
        a2.right = a5;
        a3.left = a6;
        a3.right = a7;
        TreeNode head = test.traverse(a1);
        TreeNode tail = null;
        while (head.right != null) {
            System.out.println(head.val);
            head = head.right;
        }
        System.out.println(head.val);
        tail = head;    
        System.out.println("**************");
        while (tail != null) {
            System.out.println(tail.val);
            tail = tail.left;
        }
    }
}

自己測(cè)試了下。感覺差不多了。其實(shí)就是一個(gè)簡(jiǎn)單的dfs，in order traverse this tree.
然后每次返回一個(gè)數(shù)組，第一位放head of sub tree, 第二位放tail of sub tree
然后注意雙向鏈表的連接。

Anyway, Good luck, Richardo!