<p>Given the <code>root</code> of a&nbsp;binary tree, return <em>the postorder traversal of its nodes&#39; values</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1,null,2,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,2,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" /></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1,2,3,4,5,null,8,null,null,6,7,9]</span></p>

<p><strong>Output:</strong> <span class="example-io">[4,6,7,5,2,9,8,3,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" /></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = []</span></p>

<p><strong>Output:</strong> <span class="example-io">[]</span></p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1]</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of the nodes in the tree is in the range <code>[0, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?
# use recursive dfs
# but try iterative traversal using stack(s): use a stack to insert nodes in reverse of post-order. while 
# stack is not empty, keep popping from top of stack. if top has been expanded, add to result. else, 
# expand the node and insert in reverse post-order. for each item, track if it has been expanded

# 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        stack = list()
        result = list()
        stack.append((root, False))
        while stack:
            node, expanded = stack.pop()
            if not node:    continue
            if expanded:
                result.append(node.val)
            else:
                stack.append((node, True))
                stack.append((node.right, False))
                stack.append((node.left, False))
        return result


        # def postOrder(node):
        #     if not node:    return []
        #     result = []
        #     if node.left:   result.extend(postOrder(node.left))
        #     if node.right:  result.extend(postOrder(node.right))
        #     result.append(node.val)
        #     return result

        # return postOrder(root)

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