Leetcode 114. Flatten Binary Tree to Linked List

[medium]

Given the root of a binary tree, flatten the tree into a "linked list":

• The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = 
Output: 

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

[Think]

1. need to use recursion to find the deepest left node
2. this is reverse pre-order traversal
3. create a global TreeNode variable to track current right node

[Second- DFS- using stack]

1. adding travel node on current right side, let node.left become null