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?

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# 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]