This notebook was prepared by Donne Martin. Source and license info is on GitHub.
Solution Notebook¶
Problem: Invert a binary tree.¶
Constraints¶
- What does it mean to invert a binary tree?
- Swap all left and right node pairs
- Can we assume we already have a Node class?
- Yes
- Can we assume the inputs are valid?
- No
- Can we assume this fits memory?
- Yes
Test Cases¶
Input:
5
/ \
2 7
/ \ / \
1 3 6 9
Output:
5
/ \
7 2
/ \ / \
9 6 3 1
Algorithm¶
- Base case
- If the root is None, return
- Recursive case
- Recurse on the left node
- Recurse on the right node
- Swap left and right
- Return the node
Complexity:
- Time: O(n)
- Space: O(h), where h is the height, for the recursion depth
Code¶
In [1]:
%run ../bst/bst.py
In [2]:
class InverseBst(Bst):
def invert_tree(self):
if self.root is None:
raise TypeError('root cannot be None')
return self._invert_tree(self.root)
def _invert_tree(self, root):
if root is None:
return
self._invert_tree(root.left)
self._invert_tree(root.right)
root.left, root.right = root.right, root.left
return root
Unit Test¶
In [3]:
%%writefile test_invert_tree.py
import unittest
class TestInvertTree(unittest.TestCase):
def test_invert_tree(self):
root = Node(5)
bst = InverseBst(root)
node2 = bst.insert(2)
node3 = bst.insert(3)
node1 = bst.insert(1)
node7 = bst.insert(7)
node6 = bst.insert(6)
node9 = bst.insert(9)
result = bst.invert_tree()
self.assertEqual(result, root)
self.assertEqual(result.left, node7)
self.assertEqual(result.right, node2)
self.assertEqual(result.left.left, node9)
self.assertEqual(result.left.right, node6)
self.assertEqual(result.right.left, node3)
self.assertEqual(result.right.right, node1)
print('Success: test_invert_tree')
def main():
test = TestInvertTree()
test.test_invert_tree()
if __name__ == '__main__':
main()
Overwriting test_invert_tree.py
In [4]:
%run -i test_invert_tree.py
Success: test_invert_tree