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{
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" metadata " : { } ,
" source " : [
" This notebook was prepared by [Donne Martin](https://github.com/donnemartin). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges). "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" # Challenge Notebook "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Problem: Find the lowest common ancestor in a binary tree. \n " ,
" \n " ,
" * [Constraints](#Constraints) \n " ,
" * [Test Cases](#Test-Cases) \n " ,
" * [Algorithm](#Algorithm) \n " ,
" * [Code](#Code) \n " ,
" * [Unit Test](#Unit-Test) \n " ,
" * [Solution Notebook](#Solution-Notebook) "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Constraints \n " ,
" \n " ,
" * Is this a binary search tree? \n " ,
" * No \n " ,
" * Can we assume the two nodes are in the tree? \n " ,
" * No \n " ,
" * Can we assume this fits memory? \n " ,
" * Yes "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Test Cases \n " ,
" \n " ,
" <pre> \n " ,
" _10_ \n " ,
" / \\ \n " ,
" 5 9 \n " ,
" / \\ / \\ \n " ,
" 12 3 18 20 \n " ,
" / \\ / \n " ,
" 1 8 40 \n " ,
" </pre> \n " ,
" \n " ,
" * 0, 5 -> None \n " ,
" * 5, 0 -> None \n " ,
" * 1, 8 -> 3 \n " ,
" * 12, 8 -> 5 \n " ,
" * 12, 40 -> 10 \n " ,
" * 9, 20 -> 9 \n " ,
" * 3, 5 -> 5 "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Algorithm \n " ,
" \n " ,
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" Refer to the [Solution Notebook](http://nbviewer.jupyter.org/github/donnemartin/interactive-coding-challenges/blob/master/graphs_trees/tree_lca/tree_lca_solution.ipynb). If you are stuck and need a hint, the solution notebook ' s algorithm discussion might be a good place to start. "
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]
} ,
{
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" metadata " : { } ,
" source " : [
" ## Code "
]
} ,
{
" cell_type " : " code " ,
" execution_count " : null ,
" metadata " : {
" collapsed " : true
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" source " : [
" class Node(object): \n " ,
" \n " ,
" def __init__(self, key, left=None, right=None): \n " ,
" self.key = key \n " ,
" self.left = left \n " ,
" self.right = right \n " ,
" \n " ,
" def __repr__(self): \n " ,
" return str(self.key) "
]
} ,
{
" cell_type " : " code " ,
" execution_count " : null ,
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" metadata " : { } ,
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" outputs " : [ ] ,
" source " : [
" class BinaryTree(object): \n " ,
" \n " ,
" def lca(self, root, node1, node2): \n " ,
" # TODO: Implement me \n " ,
" pass "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Unit Test "
]
} ,
{
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" metadata " : { } ,
" source " : [
" **The following unit test is expected to fail until you solve the challenge.** "
]
} ,
{
" cell_type " : " code " ,
" execution_count " : null ,
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" metadata " : { } ,
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" outputs " : [ ] ,
" source " : [
" # %lo ad test_lca.py \n " ,
" from nose.tools import assert_equal \n " ,
" \n " ,
" \n " ,
" class TestLowestCommonAncestor(object): \n " ,
" \n " ,
" def test_lca(self): \n " ,
" node10 = Node(10) \n " ,
" node5 = Node(5) \n " ,
" node12 = Node(12) \n " ,
" node3 = Node(3) \n " ,
" node1 = Node(1) \n " ,
" node8 = Node(8) \n " ,
" node9 = Node(9) \n " ,
" node18 = Node(18) \n " ,
" node20 = Node(20) \n " ,
" node40 = Node(40) \n " ,
" node3.left = node1 \n " ,
" node3.right = node8 \n " ,
" node5.left = node12 \n " ,
" node5.right = node3 \n " ,
" node20.left = node40 \n " ,
" node9.left = node18 \n " ,
" node9.right = node20 \n " ,
" node10.left = node5 \n " ,
" node10.right = node9 \n " ,
" root = node10 \n " ,
" node0 = Node(0) \n " ,
" binary_tree = BinaryTree() \n " ,
" assert_equal(binary_tree.lca(root, node0, node5), None) \n " ,
" assert_equal(binary_tree.lca(root, node5, node0), None) \n " ,
" assert_equal(binary_tree.lca(root, node1, node8), node3) \n " ,
" assert_equal(binary_tree.lca(root, node12, node8), node5) \n " ,
" assert_equal(binary_tree.lca(root, node12, node40), node10) \n " ,
" assert_equal(binary_tree.lca(root, node9, node20), node9) \n " ,
" assert_equal(binary_tree.lca(root, node3, node5), node5) \n " ,
" print( ' Success: test_lca ' ) \n " ,
" \n " ,
" \n " ,
" def main(): \n " ,
" test = TestLowestCommonAncestor() \n " ,
" test.test_lca() \n " ,
" \n " ,
" \n " ,
" if __name__ == ' __main__ ' : \n " ,
" main() "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Solution Notebook \n " ,
" \n " ,
" Review the [Solution Notebook]() for a discussion on algorithms and code solutions. "
]
}
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