您有两个问题：
1.代码中不需要if，您可以（也应该）直接在case语句上进行模式匹配。
1.在if/case的第三个子句中调用heper(Right, Acc)时，您没有保留调用helper(Left, Acc)的结果。
我没有测试它，但我认为如果您这样编写代码，您的代码应该可以工作：

helper(Tree, Acc) ->
    case Tree of
         {Value, Left, empty} -> helper(Left, Acc +1);
         {Value, empty, Right} -> helper(Right, Acc +1);
         {Value, Left, Right} -> helper(Right, helper(Left, Acc));
         _ -> Acc
    end.

事实上，您根本不需要case语句...

helper({Value, Left, empty}, Acc) -> helper(Left, Acc + 1);
helper({Value, empty, Right}, Acc) -> helper(Right, Acc + 1);
helper({Value, Left, Right}, Acc) -> helper(Right, helper(Left, Acc));
helper(_, Acc) -> Acc.

而且，如果你试图编译它，Erlang会立即告诉你Value在任何地方都是未使用的。

helper({_, Left, empty}, Acc) -> helper(Left, Acc + 1);
helper({_, empty, Right}, Acc) -> helper(Right, Acc + 1);
helper({_, Left, Right}, Acc) -> helper(Right, helper(Left, Acc));
helper(_, Acc) -> Acc.

但是，公平地说......我认为您仍然缺少{_, empty, empty}的一个子句。否则，您会将其视为具有恰好一个子节点的节点，而实际上它没有子节点......

helper({_, empty, empty}, Acc) -> Acc;
helper({_, Left, empty}, Acc) -> helper(Left, Acc + 1);
helper({_, empty, Right}, Acc) -> helper(Right, Acc + 1);
helper({_, Left, Right}, Acc) -> helper(Right, helper(Left, Acc));
helper(_, Acc) -> Acc.