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<<-DOC
A tree is considered a binary search tree (BST) if for each of its nodes the following is true:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and the right subtrees must also be binary search trees.
Removing a value x from a BST t is done in the following way:
If there is no x in t, nothing happens;
Otherwise, let t' be a subtree of t such that t'.value = x.
If t' has a left subtree, remove the rightmost node from it and put it at the root of t';
Otherwise, remove the root of t' and its right subtree becomes the new t's root.
For example, removing 4 from the following tree has no effect because there is no such value in the tree:
5
/ \
2 6
/ \ \
1 3 8
/
7
Removing 5 causes 3 (the rightmost node in left subtree) to move to the root:
3
/ \
2 6
/ \
1 8
/
7
And removing 6 after that creates the following tree:
3
/ \
2 8
/ /
1 7
You're given a binary search tree t and an array of numbers queries. Your task is to remove queries[0], queries[1],
etc., from t, step by step, following the algorithm above. Return the resulting BST.
Example
For
t = {
"value": 5,
"left": {
"value": 2,
"left": {
"value": 1,
"left": null,
"right": null
},
"right": {
"value": 3,
"left": null,
"right": null
}
},
"right": {
"value": 6,
"left": null,
"right": {
"value": 8,
"left": {
"value": 7,
"left": null,
"right": null
},
"right": null
}
}
}
and queries = [4, 5, 6], the output should be
deleteFromBST(t, queries) = {
"value": 3,
"left": {
"value": 2,
"left": {
"value": 1,
"left": null,
"right": null
},
"right": null
},
"right": {
"value": 8,
"left": {
"value": 7,
"left": null,
"right": null
},
"right": null
}
}
Input/Output
[time limit] 4000ms (rb)
[input] tree.integer t
A tree of integers.
Guaranteed constraints:
0 ≤ t size ≤ 1000,
-109 ≤ node value ≤ 109.
[input] array.integer queries
An array that contains the numbers to be deleted from t.
Guaranteed constraints:
1 ≤ queries.length ≤ 1000,
-109 ≤ queries[i] ≤ 109.
[output] tree.integer
The tree after removing all the numbers in queries, following the algorithm above.
DOC
describe "#delete_from_bst" do
<<-DOC
If there is no x in t, nothing happens;
Otherwise, let t' be a subtree of t such that t'.value = x.
If t' has a left subtree, remove the rightmost node from it and put it at the root of t';
Otherwise, remove the root of t' and its right subtree becomes the new t's root.
If t' has a left subtree and the left subtree has a child on the right,
the node you should put at the root of t' is the rightmost node (not necessarily leaf) in this subtree.
When you remove the rightmost node, you must also discard its children
(rather than keeping them as children of the rightmost node's parent).
3
/ \
2 5
/
1
remove 3:
2
/ \
1 5
remove 2:
1
\
5
remove 1
5
And removing 6 after that creates the following tree:
{ value: 3, left: { value: 2, left: { value: 1 } }, right: { value: 6, left: { value: 5 }, right: { value: 8, left: { value: 7 } } } }
3
/ \
2 6
/ / \
1 5 8
/
7
{ value: 3, left: { value: 2, left: { value: 1 } }, right: { value: 5, right: { value: 8, left: { value: 7 } } } }
3
/ \
2 5
/ \
1 8
/
7
Instead of recursively removing the largest node from the right subtree using the specified algorithm,
they want you to take the largest node's left subtree and make it the child of the largest node's parent
DOC
def remove(tree, target)
return nil if tree.nil?
if target < tree.value
tree.left = remove(tree.left, target)
elsif tree.value < target
tree.right = remove(tree.right, target)
else
if tree.left.nil?
return tree.right
elsif tree.right.nil?
return tree.left
else
max = tree.left
max = max.right while max.right
min = tree.right
min = min.left while min.left
puts [max&.value, min&.value].inspect
tree.value = max.value
tree.left = remove(tree.left, tree.value)
#tree.value = min.value
#tree.right = remove(tree.right, tree.value)
end
end
tree
end
def delete_from_bst(tree, queries)
return nil if tree.nil?
queries.each { |query| tree = remove(tree, query) }
tree
end
null = nil
[
{ t: { "value": 5, "left": { "value": 2, "left": { "value": 1, "left": null, "right": null }, "right": { "value": 3, "left": null, "right": null } }, "right": { "value": 6, "left": null, "right": { "value": 8, "left": { "value": 7, "left": null, "right": null }, "right": null } } }, queries: [4, 5, 6], x: { "value": 3, "left": { "value": 2, "left": { "value": 1, "left": null, "right": null }, "right": null }, "right": { "value": 8, "left": { "value": 7, "left": null, "right": null }, "right": null } } },
{ t: null, queries: [1, 2, 3, 5], x: nil },
{ t: { "value": 3, "left": { "value": 2, "left": null, "right": null }, "right": null }, queries: [1, 2, 3, 5], x: nil },
{ t: { "value": 3, "left": { "value": 2, "left": null, "right": null }, "right": { "value": 5, "left": null, "right": null } }, queries: [1, 2, 3, 5], x: nil },
{ t: { "value": 3, "left": { "value": 2, "left": { "value": 1, "left": null, "right": null }, "right": null }, "right": { "value": 5, "left": null, "right": null } }, queries: [3, 2, 1], x: { "value": 5, "left": null, "right": null } },
{ t: { "value": 5, "left": { "value": 3, "left": { "value": 1, "left": null, "right": null }, "right": { "value": 4, "left": null, "right": null } }, "right": { "value": 7, "left": null, "right": null } }, queries: [1, 7, 4, 6], x: { "value": 5, "left": { "value": 3, "left": null, "right": null }, "right": null } },
{ t: { value: 3, left: { value: 2, left: { value: 1 } }, right: { value: 6, left: { value: 5 }, right: { value: 8, left: { value: 7 } } } }, queries: [6], x: { value: 3, left: { value: 2, left: { value: 1 } }, right: { value: 5, right: { value: 8, left: { value: 7 } } } } },
{ t: { "value": 3, "left": { "value": 2, "left": { "value": 1, "left": null, "right": null }, "right": null }, "right": { "value": 5, "left": null, "right": null } }, queries: [3], x: { "value": 2, "left": { "value": 1, "left": null, "right": null }, "right": { "value": 5, "left": null, "right": null } } },
{ t: null, queries: [100000000, -1000000000, 0, 1, -1, 2, -2], x: null },
{ t: { "value": 3, "left": { "value": 2, "left": null, "right": null }, "right": { "value": 5, "left": null, "right": null } }, queries: [2, 3, 0, 5], x: nil },
].each do |x|
it do
result = delete_from_bst(Tree.build_from(x[:t]), x[:queries])
expected = x[:x] ? Tree.build_from(x[:x]).to_s : nil
expect(result ? result.to_s : result).to eql(expected)
end
end
it do
require_relative 'spec_8'
x = SPEC8
result = delete_from_bst(Tree.build_from(x[:t]), x[:queries])
expected = x[:x] ? Tree.build_from(x[:x]).to_s : nil
expect(result ? result.to_s : result).to eql(expected)
end
it do
require_relative 'spec_9'
x = SPEC9
result = delete_from_bst(Tree.build_from(x[:t]), x[:queries])
expected = x[:x] ? Tree.build_from(x[:x]).to_s : nil
expect(result ? result.to_s : result).to eql(expected)
end
it do
require_relative 'spec_10'
x = SPEC10
result = delete_from_bst(Tree.build_from(x[:t]), x[:queries])
expected = x[:x] ? Tree.build_from(x[:x]).to_s : nil
expect(result ? result.to_s : result).to eql(expected)
end
end
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