Am I correct in saying that
traverse(node): if node is null, return print node traverse(node's right subtree) traverse(node's left subtree)
would produce output that is the reverse of post-order traversal?
post-order(node): if node is null, return post-order(node's left subtree) post-order(node's right subtree) print node
I am mostly interested because if this is true, it greatly simplifies the iterative method for post-order traversal. It "feels" right with a hand-wavey explanation - and with testing on some trees - but how would I prove it?