Given the root node root of a binary tree, return the values of its nodes in postorder traversal order.

In postorder traversal, you visit the left subtree, then the right subtree, and finally the current node.

Note: A recursive approach is straightforward. Try to solve it using an iterative method.

Constraints:

• The number of nodes in the tree is in the range $[0, 100]$.

• $-100 \leq$ Node.val $\leq 100$

Input: root = [1, null, 2, 3]-TreeNode
Output: [3, 2, 1]-List[int]
Explanation: Node 1 has no left child and right child 2, and node 2 has left child 3. Postorder visits the left subtree (3), then the right subtree (2’s right is null), then node 2, and finally node 1, giving [3, 2, 1].

Input: root = []-TreeNode
Output: []-List[int]
Explanation: An empty tree has no nodes to visit, so the postorder traversal is empty.

Input: root = [1]-TreeNode
Output: [1]-List[int]
Explanation: With a single node, postorder traversal visits that node last (and also first), resulting in [1].

Input: root = [1, 2, 3, 4, 5, null, 6]-TreeNode
Output: [4, 5, 2, 6, 3, 1]-List[int]
Explanation: Root 1 has left child 2 and right child 3; node 2 has children 4 and 5, and node 3 has right child 6. Postorder visits the left subtree of 1 ([4, 5, 2]), then the right subtree of 1 ([6, 3]), then node 1, giving [4, 5, 2, 6, 3, 1].

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