"""

Given the root of a binary tree, return the inorder traversal of its nodes' values.
 
Example 1:


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

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:


Input: root = [1,2]
Output: [2,1]

Example 5:


Input: root = [1,null,2]
Output: [1,2]

 
Constraints:

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

 
Follow up:
Recursive solution is trivial, could you do it iteratively?
 

"""


# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        def inorder(node, ls):
            if n
     
        ls = []
        inorder(root, ls)
        return ls
ode is None:
                return
            inorder(node.left, ls)
            ls.append(node.val)
            inorder(node.right, ls)
   