题目
Convert a BST to a sorted circular doubly-linked list in-place. Think of the left and right pointers as synonymous to the previous and next pointers in a doubly-linked list.

答案
很暴力的思路，很多题都可以用, 不过不符合in-place的要求

class Solution {    
    public Node treeToDoublyList(Node root) {
        if(root == null) return null;
        // Use an array list to store all treenodes in sorted order
        List<Node> list = new ArrayList<>();
        recur(root, list);
        
        for(int i = 0; i < list.size() - 1; i++) {
            Node curr = list.get(i);
            Node next = list.get(i + 1);
            curr.right = next;
            next.left = curr;
        }
        
        Node first = list.get(0);
        Node last = list.get(list.size() - 1);
        first.left = last;
        last.right = first;
        
        return first;
    }
    
    public void recur(Node root, List<Node> list) {
        if(root == null) return;
        recur(root.left, list);
        list.add(root);
        recur(root.right, list);
    }
}

这个思路不难想，但是需要用global var
基本就是in order traversal过一遍BST,在途中连接上一个访问的结点和当前的结点

class Solution {
    Node zero = null;
    Node curr = null;
    public Node treeToDoublyList(Node root) {
        if(root == null) return null;
        Node zero = new Node(0);
        curr = zero;

        recur(root);   
        curr.right = zero.right;
        zero.right.left = curr;
        
        return zero.right;
    }
    
    public void recur(Node root) {
        if(root == null) return;
        recur(root.left);
        
        curr.right = root;
        root.left = curr;
        curr = root;
        
        recur(root.right);
    }
}