题目
Convert a Binary Search Tree to a sorted Circular Doubly-Linked List in place.
You can think of the left and right pointers as synonymous to the predecessor and successor pointers in a doubly-linked list. For a circular doubly linked list, the predecessor of the first element is the last element, and the successor of the last element is the first element.
We want to do the transformation in place. After the transformation, the left pointer of the tree node should point to its predecessor, and the right pointer should point to its successor. You should return the pointer to the smallest element of the linked list.
解题思路
DFS中序遍历，记录最小的节点（第一个节点）和上一个节点，当前节点左指针指向上一个节点，上一个节点的右指针指向当前节点，最后再把第一个和最后一个节点连接起来
代码
class Solution:
first, last = None, None
def treeToDoublyList(self, root: 'Optional[Node]') -> 'Optional[Node]':
def dfs(root):
if not root:
return
dfs(root.left)
if self.last:
self.last.right = root
root.left = self.last
else:
self.first = root
self.last = root
dfs(root.right)
if not root:
return None
dfs(root)
self.last.right = self.first
self.first.left = self.last
return self.first
