Python Checking paths to leaf in binary tree python giving data in the leaf

Question

Lets say i have this tree:

                                     cough
                      Yes /                            \\ No
                   sneezing            


        

Answer 1

Here's what I came up with

class Tree:
  def __init__(self, data, left=None, right=None):
    self.data = data
    self.left = left
    self.right = right

  @property
  def is_leaf(self):
    return not (self.left or self.right)

  def __repr__(self):
    return 'Tree({}, {}, {})'.format(self.data, self.left, self.right)

  def find(self, target, path_to=()):
    if self.is_leaf:
      if self.data == target:
        yield path_to
    else:
      if self.left:
        yield from self.left.find(target, (*path_to, True))
      if self.right:
        yield from self.right.find(target, (*path_to, False))

t = Tree('Cough', Tree('Sneezing', Tree('Fever', Tree('Dead'), Tree('Cold')), Tree('Fever', Tree('Influenza'), Tree('Cold'))), Tree('Sneezing', Tree('Fever', Tree('Dead'), Tree('Influenza')), Tree('Fever', Tree('Cold'), Tree('Healthy'))))

print(list(t.find('Influenza')))

By having our find method be a generator, we can easily bubble positive results up the call stack using yield from. If you're using a version of Python that doesn't support argument unpacking (*path_to, True), then path_to + (True,) is equivalent

Edit: Here's a version the doesn't use yield

def find(self, target, path_to=()):
  if self.is_leaf:
    if self.data == target:
      return [path_to]
    else:
      return []
  else:
    if self.left:
      l = self.left.find(target, (*path_to, True))
    if self.right:
      r = self.right.find(target, (*path_to, False))
    return l + r