week 6

This week we begin with binary trees. Its supreme node is called the root. And especially for binary trees, each node can only have 2 children. In the list expression, recursion is obvious in that the tl[0] is an object/ value while the rest two component are tree lists themselves. As for the three methods to trace the tree, I find the first one is what we generally use, while i really don't know what the other two can be used in. Binary tree presents a systematic hierarchy of data structure.
In addiction, i come up with a different solution from class: e.g. in order [tl[0]] + (inorder(tl[1]) if isinstance(tl[1], list) else []) + (inorder(tl[2]) if isinsatnce(tl[2], list) else []). The ternary structure requires all three components. And this solution is just a thought, I later find out this would not work if the list is None.

The second part is the general tree implementation, where the trees can have more than two children. It follow the structure from binary tree except that it combines the branches of each nodes together in a list(from left to right). Clearly, the leaves are also presented as subtrees with children set to empty list in its __init__ method. I spent a lot of time trying to figure out how to write its methods in recursion and find that though it's relatively easy to trace and understand recursion; it's so much harder to write them down on my own. By looking at the answers, i gather some inspiration from then.
__contain__: I was initially kind of confused about that it doesn't  have a base case. But when self.children is [], the for loop wouldn't executed, thus ending the recursion.  Notably, the use of "any" is helpful in some cases
__repr__: I was trying this in the shell, and it turned out that i don't need to use recursion at all. Since i pass in a tree list, the __init__ method__ would turn it into the appropriate structure, and when we tried to "represent" it. Can't we just simply use "Tree()". format(self.value, self.children)?
leaf_count: this function also used something new that took me a while to understand. The idea I frist come up with is sum([leaf_count(i) ) for i in t.children),  then i am stuck with the base case.
count: the base case is when self.children == [], return 1 e.g.sum([2**x for x in []]) return 0

Meanwhile, I am trying to come up with an ultimate process to write recursion function. The first step is coming up with the base case. And write code that recursion can run.  But loopholes are everywhere.

Questions
1. if statement can exist without else, in the short-form list expression can we use ### if ## and leave out the else?  when i tried this, it results in an error.
answer: [x+7 if x> 2 else x+2 for x in [2,3,4]] ternary fixed syntax

2. when we define our own class, isinstance method should also work on defined classes. But for the Tree class here, even though we require children to be None or another Treelist, isinstance can't tell the difference. AKA isinstance(Tree(2, [2,3]), Tree) evaluates to True. How can we fix the problem?