This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Challenge Notebook

Problem: Determine if a tree is a valid binary search tree.

Constraints

  • Can the tree have duplicates?
    • Yes
  • If this is called on a None input, should we raise an exception?
    • Yes
  • Can we assume we already have a Node class?
    • Yes
  • Can we assume this fits in memory?
    • Yes

Test Cases

Valid:
      5
    /   \
   5     8
  /     /
 4     6
        \
         7

Invalid:
      5
    /   \
   5     8
  / \   /
 4   9 7

Algorithm

Refer to the Solution Notebook. If you are stuck and need a hint, the solution notebook's algorithm discussion might be a good place to start.

Code

In [ ]:
%run ../bst/bst.py
%load ../bst/bst.py
In [ ]:
class BstValidate(Bst):

    def validate(self):
        # TODO: Implement me
        pass

Unit Test

The following unit test is expected to fail until you solve the challenge.

In [ ]:
# %load test_bst_validate.py
import unittest


class TestBstValidate(unittest.TestCase):

    def test_bst_validate_empty(self):
        bst = BstValidate(None)
        bst.validate()

    def test_bst_validate(self):
        bst = BstValidate(Node(5))
        bst.insert(8)
        bst.insert(5)
        bst.insert(6)
        bst.insert(4)
        bst.insert(7)
        self.assertEqual(bst.validate(), True)

        bst = BstValidate(Node(5))
        left = Node(5)
        right = Node(8)
        invalid = Node(20)
        bst.root.left = left
        bst.root.right = right
        bst.root.left.right = invalid
        self.assertEqual(bst.validate(), False)

        print('Success: test_bst_validate')


def main():
    test = TestBstValidate()
    test.assertRaises(TypeError, test.test_bst_validate_empty)
    test.test_bst_validate()


if __name__ == '__main__':
    main()

Solution Notebook

Review the Solution Notebook for a discussion on algorithms and code solutions.