If we totally reverse the post-order traverse (new_traverse = postorder[::-1]), it becomes a 'pre-order' traverse that visit father->right child-> left child
Similarly, if we reverse in-order traverse, it becomes a 'in-order' traverse that visit right->father->left.
Then, these two new traverse is the same as the traverse in question 105. The only difference is that left becomes right and right becomes left. We can use same method in 105 to solve it. Below is a sample of iterative solution. Hope this helps.
class Solution(object): def buildTree(self, inorder, postorder): """ :type inorder: List[int] :type postorder: List[int] :rtype: TreeNode """ inorder = inorder[::-1] preorder = postorder[::-1] if preorder == : return None d1,d2 = 0,0 head = TreeNode(preorder) stack = [head] while d1 < len(preorder)-1: node = stack.pop() if node.val != inorder[d2]: stack.append(node) new_node = TreeNode(preorder[d1+1]) node.right = new_node stack.append(new_node) d1 += 1 else: while stack !=  and stack[-1].val == inorder[d2+1]: node = stack.pop() d2 += 1 d2 +=1 d1 +=1 new_node = TreeNode(preorder[d1]) node.left = new_node stack.append(new_node) return head