# 145 Binary Tree Postorder Traversal

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

Example 1:

`````` 1
\
2
/
3

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

Example 2:

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

Example 3:

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

Example 4:

``````   1
/
2

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

Example 5:

`````` 1
\
2

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

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?

 `````` 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 `````` ``````# 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]: def postorder(node: Optional[TreeNode], lst: List[int]): if not node: return postorder(node.left, lst) postorder(node.right, lst) lst.append(node.val) result = [] postorder(root, result) return result ''' Iterative ''' class Solution: def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]: if root is None: return [] stack, res = [root], [] while stack: node = stack.pop() if node: # pre-order, right first res.append(node.val) stack.append(node.left) stack.append(node.right) # reverse result return res[::-1]``````