给定一个二叉树的根节点 root ，返回它的 中序 遍历。
示例 1：
输入：root = [1,null,2,3]
输出：[1,3,2]
示例 2：
输入：root = []
输出：[]
示例 3：
输入：root = [1]
输出：[1]
示例 4：
输入：root = [1,2]
输出：[2,1]
示例 5：
输入：root = [1,null,2]
输出：[1,2]
提示：
[0, 100] 内-100 <= Node.val <= 100
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
## template ```python class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class List2Tree(object): def __init__(self, nums: list): self.nums = nums self.queue = [] if len(nums) == 1: self.root = TreeNode(self.nums.pop(0)) else: a = self.nums.pop(0) b = self.nums.pop(0) c = self.nums.pop(0) self.root = TreeNode(a) if b is not None: self.root.left = TreeNode(b) else: self.root.left = b if c is not None: self.root.right = TreeNode(c) else: self.root.right = c self.queue.append(self.root.left) self.queue.append(self.root.right) def convert(self): while len(self.nums) > 0 and len(self.queue)> 0: node = self.queue.pop(0) if node is not None: num= self.nums.pop(0) if num is not None: node.left = TreeNode(num) else: node.left = num if len(self.nums) > 0: num = self.nums.pop(0) else: num = None if num is not None: node.right = TreeNode(num) else: node.right = num self.queue.append(node.left) self.queue.append(node.right) return self.root class Solution(object): def inorderTraversal(self, root): if root is None: return [] root = List2Tree(root).convert() res = [] stack = [root] while len(stack) > 0: curr = stack.pop() if not isinstance(curr, TreeNode): res.append(curr) continue if curr.right is not None: stack.append(curr.right) stack.append(curr.val) if curr.left is not None: stack.append(curr.left) return res # %% s = Solution() print(s.inorderTraversal(root = [1,None,2,3])) ``` ## 答案 ```python ``` ## 选项 ### A ```python ``` ### B ```python ``` ### C ```python ```