# 94 Binary Tree Inorder Traversal

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

Example 1:

`````` 1
\
2
/
3

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

Example 2:

``````Input: root = []
Output: []
``````

Example 3:

``````Input: root = 
Output: 
``````

Example 4:

``````   1
/
2

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

Example 5:

`````` 1
\
2

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?

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 `````` ``````# 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 ''' Recursive ''' class Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: def inorder(node: Optional[TreeNode], lst: List[int]): if not node: return inorder(node.left, lst) lst.append(node.val) inorder(node.right, lst) result = [] inorder(root, result) return result ''' Iterative ''' class Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: if root is None: return [] stack, res = [], [] curr = root while curr or stack: while curr: stack.append(curr) curr = curr.left curr = stack.pop() res.append(curr.val) # Add after all left children curr = curr.right return res``````