<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?
# 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

# post order traversal: Left Right Root
# 4 6 7 5 2 9 8 3 1
# pre order traversal: Root Right Left
# 1 3 8 9 2 5 7 6 4

# postorder can be implemented as preorder traversal but by visiting right child first and then left child
# pre order traversal: Root Left Right but for this scenario we can traverse as Root Right Left

class Solution:

    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if root == None:
            return []
        stack = [root]
        result = []

        while stack : 
            node = stack.pop()
            result.append(node.val)

            if node.left:
                stack.append(node.left)

            if node.right:
                stack.append(node.right)

        return result[::-1]
        
        
16100