in-order traversal of a binary search tree returning a list

Question

I need to return an ordered list of references with an empty in between them after traversing a BST, but i can't workout how to do it.

def inorder(self, Node):

    res = ""
    if Node != None:

        self.inorder(Node.LeftChild())
        res = res +(Node.key + " " + Node.payload + "\n")
        self.inorder(Node.RightChild())

    return res

this is my code, so far i can only get one reference returned, any help will be much appreciated.

you process left branch and throw the result away. Same for right branch. Your method has a return value, use it. — Lesiak
– Lesiak, Commented Mar 19, 2020 at 22:29
thanks @Lesiak, i'm new at this, won't you be able to provide an example how to do it? — eLlobregat
– eLlobregat, Commented Mar 19, 2020 at 22:34
Variable and function names should follow the lower_case_with_underscores style. — AMC
– AMC, Commented Mar 19, 2020 at 23:07

kareem_emad · Accepted Answer · 2020-03-19 22:47:52Z

1

You almost had it right, just need to use the return value of the right and the left sub trees in your res variable

def inorder(self, Node):

    res = ""
    if Node != None:

        lres = self.inorder(Node.LeftChild())
        rres = self.inorder(Node.RightChild())
        res = lres + (Node.key + " " + Node.payload + "\n") + rres

    return res

Note: res was always empty string in your condition, so you did not need to do res = res+, also there is a shorthand notation for doing this res+= but you didn't need it anyway.

edited Mar 19, 2020 at 22:47

answered Mar 19, 2020 at 22:42

kareem_emad

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