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.
Input: root = [4,2,5,1,3]
Output: [1,2,3,4,5] Explanation:
The figure below shows the transformed BST. The solid line indicates the successor relationship, while the dashed line means the predecessor relationship.
Input: root = [2,1,3] Output: [1,2,3]
Input: root =  Output:  Explanation: Input is an empty tree. Output is also an empty Linked List.
Input: root =  Output: 
Find the BST tree values in ascending order. Then use the BST tree values to build a Circular Doubly-Linked List.
""" # Definition for a Node. class Node: def __init__(self, val, left=None, right=None): self.val = val self.left = left self.right = right """ class Solution: def treeToDoublyList(self, root: 'Node') -> 'Node': if not root: return None node_values =  self.get_node_values(root, node_values) first_node = Node(node_values) prev = first_node for value in node_values[1:]: new_node = Node(value) new_node.left = prev prev.right = new_node prev = new_node prev.right = first_node first_node.left = prev return first_node def get_node_values(self, root, results): if not root: return self.get_node_values(root.left, results) results.append(root.val) self.get_node_values(root.right, results)
- Time Complexity: O(N).
- Space Complexity: O(N).