## Question

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]

**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?

## Python Solution

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: stack = [] l = [] curr = root while 1: if curr is not None: stack.append(curr) curr = curr.left elif stack: x = stack.pop() l.append(x.val) curr = x.right else: break return l