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<<-DOC
Note: Your solution should have only one BST traversal and O(1) extra space complexity,
since this is what you will be asked to accomplish in an interview.
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.
Given a binary search tree t, find the kth largest element in it.
Note that kth largest element means kth element in increasing order. See examples for better understanding.
Example
For
t = {
"value": 3,
"left": {
"value": 1,
"left": null,
"right": null
},
"right": {
"value": 5,
"left": {
"value": 4,
"left": null,
"right": null
},
"right": {
"value": 6,
"left": null,
"right": null
}
}
}
and k = 2, the output should be
kthLargestInBST(t, k) = 3.
Here is what t looks like:
3
/ \
1 5
/ \
4 6
The values of t are [1, 3, 4, 5, 6], and the 2nd element when the values are in sorted order is 3.
For
t = {
"value": 1,
"left": {
"value": -1,
"left": {
"value": -2,
"left": null,
"right": null
},
"right": {
"value": 0,
"left": null,
"right": null
}
},
"right": null
}
and k = 1, the output should be
kthLargestInBST(t, k) = -2.
Here is what t looks like:
1
/
-1
/ \
-2 0
The values of t are [-2, -1, 0, 1], and the 1st largest is -2.
Input/Output
[time limit] 4000ms (rb)
[input] tree.integer t
A tree of integers. It is guaranteed that t is a BST.
Guaranteed constraints:
1 ≤ tree size ≤ 104,
-105 ≤ node value ≤ 105.
[input] integer k
An integer.
Guaranteed constraints:
1 ≤ k ≤ tree size.
[output] integer
The kth largest value in t
DOC
describe "#kth_largest_in_bst" do
def kth_largest_in_bst(tree, k)
puts tree.to_a.inspect
stack = [tree]
all = [ ]
until stack.empty?
node = stack.pop
all.push(node.value)
stack.push(node.left) if node.left
stack.push(node.right) if node.right
end
puts all.inspect
all[k + 1]
end
def to_array(tree)
return [] if tree.nil?
to_array(tree.left) + [tree.value] + to_array(tree.right)
end
def kth_largest_in_bst(tree, k)
to_array(tree)[k - 1]
end
class Traversal
attr_reader :k
def initialize(k)
@counter = 0
@k = k
end
def traverse(tree)
return if tree.nil?
if @counter < k && result = traverse(tree.left)
return result
end
@counter += 1
return tree.value if @counter == k
traverse(tree.right) if @counter < k
end
end
def kth_largest_in_bst(tree, k)
Traversal.new(k).traverse(tree)
end
def traverse(node, yielder)
return if node.nil?
traverse(node.left, yielder)
yielder.yield node.value
traverse(node.right, yielder)
end
def kth_largest_in_bst(tree, k)
x = Enumerator.new { |yielder| traverse(tree, yielder) }
(k-1).times { x.next }
x.next
end
[
{ t: { "value": 3, "left": { "value": 1, "left": nil, "right": nil }, "right": { "value": 5, "left": { "value": 4, "left": nil, "right": nil }, "right": { "value": 6, "left": nil, "right": nil } } }, k: 2, x: 3 },
{ t: { "value": 1, "left": { "value": -1, "left": { "value": -2, "left": nil, "right": nil }, "right": { "value": 0, "left": nil, "right": nil } }, "right": nil }, k: 1, x: -2 },
{ t: { "value": 100, "left": nil, "right": nil }, k: 1, x: 100 },
{ t: { "value": 1, "left": { "value": 0, "left": nil, "right": nil }, "right": { "value": 2, "left": nil, "right": nil } }, k: 3, x: 2 },
{ t: { "value": 1, "left": { "value": 0, "left": nil, "right": nil }, "right": { "value": 2, "left": nil, "right": nil } }, k: 2, x: 1 },
{ t: { "value": -2, "left": nil, "right": { "value": 3, "left": { "value": 2, "left": nil, "right": nil }, "right": nil } }, k: 1, x: -2 }
].each do |x|
it do
expect(kth_largest_in_bst(Tree.build_from(x[:t]), x[:k])).to eql(x[:x])
end
end
end
|
