Python Binary Search Tree Implementation

Question

#BST
class Node:
    def __init__(self, val, left = None, right = None):
        self.val = val
        self.left = left
        self.right = right

def search(root, k):
    if root is None or root.val == k:
        return root
    elif root.val < k:
        return search(root.right, k)
    else:
        return search(root.left, k)

def minimum(root):
    if root is None or root.left is None:
        return root
    return minimum(root.left)

def maximum(root):
    if root is None or root.right is None:
        return root
    return maximum(root.right)

def successor(root, node):
    if node.right is not None:
        return minimum(node.right)
    succ = None
    while root is not None and root.val != node.val:
        if root.val > node.val:
            succ = root
            root = root.left
        else:
            root = root.right
    return succ

def predecessor(root, node):
    if node.left is not None:
        return maximum(node.left)
    else:
        pred = None
        while root is not None and root.val != node.val:
            if root.val > node.val:
                root = root.left
            else:
                pred = root
                root = root.right
        return pred

def insert(root, key):
    if root is None:
        return Node(key)
    elif root.val < key:
        return Node(root.val, root.left, insert(root.right, key))
    else:
        return Node(root.val, insert(root.left, key), root.right)

def parent(root, node):
    if root is None:
        return root
    elif (root.left is not None and root.left.val == node.val) or \
         (root.right is not None and root.right.val == node.val):
        return root
    else:
        if root.val < node.val:
            return parent(root.right, node)
        else:
            return parent(root.left, node)

def delete(root, node):
    p = parent(root, node)
    if node.left is None or node.right is None:
        replacement = node.left if node.left is None else node.right
        if p is None:
            root = replacement
        elif p.val < node.val:
            p.right = replacement
        else:
            p.left = replacement
    else:
        replacement = successor(root, node)
        succ_parent = parent(root, replacement)
        succ_parent.left = replacement.right
        node.val = replacement.val
    return root

Require code review in terms of correctness, style consistency, efficiency, code readability, and "pythonic"ness. (I, of course, welcome any other comments on the code as well)

I know that this a bit big to review completely. Reviewing only one or more methods will be appreciated.

Also, the WikiPage talks about designing persistent data structures. When should one consider such design and what are the advantages?

Github: https://github.com/abhisheksp/algorithms (the repo has the test cases)

short gripe: if not None is very redundant. None type in python has a truthy value of false. You can jut say if root.right: and the code will then execute if it exists, and will not if it doesn't. This alone will make your code much more readable. — Erich
– Erich, Commented Aug 30, 2017 at 22:22
@Erich I feel that explicit "is None" comparison is more readable. Even the PEP style guide recommends that. Source: docs.quantifiedcode.com/python-anti-patterns/readability/… — Abhishek
– Abhishek, Commented Aug 30, 2017 at 22:48
Is there a reason you can't save the parent of a node and update it as it changes? — Erich
– Erich, Commented Aug 31, 2017 at 0:04

J_H · Accepted Answer · 2017-08-31 00:04:13Z

docstrings

Docstrings matter. Delete the cryptic #BST comment and spell it out in a PEP8 class docstring:

class Node:
"""Implements a Binary Search Tree."""

Either search or __init__ needs a docstring that mentions the equality case (I'm looking at elif root.val < k:, as opposed to <= k).

correctness

I can't imagine successor is correct. You didn't offer any unit tests like this, but I'm thinking of passing in a node with no right subtree, a node having the same value as root. Then we return None ?? This comes back to the <= k detail I mentioned above. Anyway, predecessor is nicely parallel, simple is good. And insert is beautiful and concise.

public API design

For some methods, rather than passing in root, it would be more natural to be working with an instance of your BST class, with self.root implicitly available.

style

As a trivial style detail, in parent you might use parentheses a level deeper to remove the backslash.

I didn't run flake8 myself, but overall it looks good. I have a weak preference for if node.left: instead of if node.left is not None:.

tests

I'm looking at this:

class BSTTest(unittest.TestCase):
    def test_recursive_search(self):
        root  = build_bst()
        actual = search(root, 35).val
        expected = 35
        self.assertEqual(expected, actual)

I would much rather see the method expressed concisely:

    def test_recursive_search(self):
        root  = build_bst()
        self.assertEqual(35, search(root, 35).val)

It might even come out as:

    def test_recursive_search(self):
        self.assertEqual(35, search(self.root, 35).val)

where build_bst() is invoked by a setup method.

There are other tests, e.g. test_insert_on_empty_tree, where of course hanging on to an actual temp variable is the natural way to express it.

I am very glad to see these tests. They appear to give reasonable code coverage.

Another example tree that stores multiple copies of a given value (e.g. 30) would be of interest.

summary

Overall it looks good! Ship it.

chatton · Accepted Answer · 2017-08-31 15:04:11Z

Thanks for sharing your code,

Instead of dealing with nodes directly, I think it would be easier, and make a lot more sense to have code that looked like this.

bst = BST([1,6,3,3,5])
bst.insert(5)
min = bst.minimum()
max = bst.max()

etc.

the BST class would be in charge of managing the Node objects. This way I'm not passing around a root Node to all my functions. For someone who wants to use this tree, I shouldn't need to know about how all these Nodes work. A Node is an implementation detail of your BST

Make use of __eq__, __hash__and __str__. Implementing these methods would allow you to compare and display your Nodeobjects very easily. You could also implement __lt__ and __gt__ to compare nodes via node1 < node2 rather than node1.val < node2.val (although __gt__ may not actually be required here)

very small stylistic thing here. The code left = None, right = None should be left=None, right=None (no spaces) according to PEP8 guidelines.

Some additional methods you could add

post_order_traversal
pre_order_traversal
in_order_traversal

these could maybe just print the elements in the given order, or maybe return a list of the elements (not the nodes) in that order.

Overall I think it's looking very nice!

Keep it up

stefan · Accepted Answer · 2017-09-01 12:28:50Z

As you asked for persistence.

You can temporarily instrument your code like

def __init__(self, val, left = None, right = None):
    print("new", self.val, self)
    ...

def __del__(self):
    print("del", self.val, self)

to get an idea what is going on whe you are inserting nodes. Your tree is not persistent as nodes are deleted and recreated with the same value.

r = Node(0)
for i in range(4):
    r=insert(r,i+1)

gives

new 0 <Node object at 0x7f6dc2d1b240>
new 1 <Node object at 0x7f6dc2d2e2e8>
new 0 <Node object at 0x7f6dc2d2e390>
del 0 <Node object at 0x7f6dc2d1b240>
new 2 <Node object at 0x7f6dc2d1b240>
new 1 <Node object at 0x7f6dc2d2e3c8>
new 0 <Node object at 0x7f6dc2d2e358>
del 0 <Node object at 0x7f6dc2d2e390>
del 1 <Node object at 0x7f6dc2d2e2e8>
new 3 <Node object at 0x7f6dc2d2e390>
new 2 <Node object at 0x7f6dc2d2e320>
new 1 <Node object at 0x7f6dc2d2e470>
new 0 <Node object at 0x7f6dc2d2e2b0>
del 0 <Node object at 0x7f6dc2d2e358>
del 1 <Node object at 0x7f6dc2d2e3c8>
del 2 <Node object at 0x7f6dc2d1b240>

so any other variable referencing a node or even the root node will possibly reference something old.

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Python Binary Search Tree Implementation

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