'None' in python

Question

I have a simple program above that creates a BST from a sorted array. It should parse the tree without showing the leaves which are essentially None. Could someone help explain why the program still spits out 'None'. I'm a python NooB and would appreciate any help.I have tried != 'None' along with is None but get the same results.

class Node:
    def __init__(self,value):
        self.value=value
        self.nodeleft=None
        self.noderight=None

def makeBST(ia,start,end,tree):
    if (end < start):
        return None
    mid = (start + end) / 2
    n = Node(ia[mid])
    n.nodeleft = makeBST(ia, start, mid-1, tree)
    n.noderight = makeBST(ia, mid+1, end, tree)
    tree.append(n)
    return n

def printBST(root):
    print 'RR' ,root.value
    if root.nodeleft == None:
        print 'EOT'
    else:    
        print printBST(root.nodeleft)

    if root.noderight == None:
        print 'EOT'
    else:    
        print printBST(root.noderight)

if __name__ == '__main__':
    array = [1, 2, 3, 4, 5, 6]
    dic = []
    root = makeBST(array, 0, len(array)-1, dic)
    printBST(root)

David Heffernan · Accepted Answer · 2013-10-06 13:41:21Z

The problem is that your code was passing the return value of printBST to print. Since printBST does not return anything, None was printed.

So when you wrote:

print printBST(root.nodeleft)

that code is certain to print None because printBST does not contain a return statement and so defaults to returning None.

You need to change printBST to do this:

def printBST(root):
    print 'RR' ,root.value
    if root.nodeleft is None:
        print 'EOT'
    else:    
        printBST(root.nodeleft)

    if root.noderight is None:
        print 'EOT'
    else:    
        printBST(root.noderight)

Note also that using is is the correct way to test for None.

That said, you can make your code simpler like this:

def printBST(root):
    if root is None:
        print 'EOT'
        return
    print 'RR', root.value
    printBST(root.nodeleft)
    printBST(root.noderight)

As well as being simpler, this code has the additional benefit of not failing when presented with an empty tree.

TerryA · Accepted Answer · 2013-10-06 13:46:37Z

2

printBST should return the values, and not print them. Because it does not return anything, it defaults to None. That is why printBST(root) is None

printBST(root) by itself won't print the value on its own. You have to put a print before:

print printBST(root)

Per PEP 8, you should never compare the NoneType singleton with equality operators (eg == and !=). Use is None and/or is not None

edited Oct 6, 2013 at 13:46

answered Oct 6, 2013 at 13:30

TerryA

60.2k11 gold badges122 silver badges148 bronze badges

1 Comment

David Heffernan Over a year ago

I did not downvote, but the answer doesn't make much sense. It looks awfully like the design of printBST is indeed that it should print the tree. It's not difficult to fix it to work as desired. And printBST(root) is indeed None, but then that is never printed. What is printed is printBST(root.nodeleft) and printBST(root.noderight). You should be asking why you are getting upvotes! ;-)

Collectives™ on Stack Overflow

'None' in python

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related