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<<-DOC
Given a binary tree t and an integer s,
determine whether there is a root to leaf path in t
such that the sum of vertex values equals s.
Example
For
t = {
"value": 4,
"left": {
"value": 1,
"left": {
"value": -2,
"left": null,
"right": {
"value": 3,
"left": null,
"right": null
}
},
"right": null
},
"right": {
"value": 3,
"left": {
"value": 1,
"left": null,
"right": null
},
"right": {
"value": 2,
"left": {
"value": -2,
"left": null,
"right": null
},
"right": {
"value": -3,
"left": null,
"right": null
}
}
}
}
and
s = 7,
the output should be hasPathWithGivenSum(t, s) = true.
This is what this tree looks like:
4
/ \
1 3
/ / \
-2 1 2
\ / \
3 -2 -3
Path 4 -> 3 -> 2 -> -2 gives us 7, the required sum.
For
t = {
"value": 4,
"left": {
"value": 1,
"left": {
"value": -2,
"left": null,
"right": {
"value": 3,
"left": null,
"right": null
}
},
"right": null
},
"right": {
"value": 3,
"left": {
"value": 1,
"left": null,
"right": null
},
"right": {
"value": 2,
"left": {
"value": -4,
"left": null,
"right": null
},
"right": {
"value": -3,
"left": null,
"right": null
}
}
}
}
and
s = 7,
the output should be hasPathWithGivenSum(t, s) = false.
This is what this tree looks like:
4
/ \
1 3
/ / \
-2 1 2
\ / \
3 -4 -3
There is no path from root to leaf with the given sum 7.
Input/Output
[time limit] 4000ms (rb)
[input] tree.integer t
A binary tree of integers.
Guaranteed constraints:
0 ≤ tree size ≤ 5 · 104,
-1000 ≤ node value ≤ 1000.
[input] integer s
An integer.
Guaranteed constraints:
-4000 ≤ s ≤ 4000.
[output] boolean
Return true if there is a path from root to leaf in t such that the sum of node values in it is equal to s,
otherwise return false.
DOC
describe "#path_with_given_sum?" do
def leaf?(node)
node.left.nil? && node.right.nil?
end
def path_with_given_sum?(tree, target_sum)
stack = []
stack.push(tree) if tree
sum = 0
until stack.empty?
node = stack.pop
sum += node.value
return true if leaf?(node) && sum == target_sum
stack.push(node.left) if node.left
stack.push(node.right) if node.right
end
sum == target_sum
end
def path_with_given_sum?(tree, target)
return target.zero? if tree.nil?
new_target = target - tree.value
return new_target.zero? if leaf?(tree)
return true if tree.left && path_with_given_sum?(tree.left, new_target)
return true if tree.right && path_with_given_sum?(tree.right, new_target)
false
end
def path_with_given_sum?(tree, target)
return target.zero? if tree.nil?
new_target = target - tree.value
path_with_given_sum?(tree.left, new_target) || path_with_given_sum?(tree.right, new_target)
end
[
{ t: { "value": 4, "left": { "value": 1, "left": { "value": -2, "left": nil, "right": { "value": 3, "left": nil, "right": nil } }, "right": nil }, "right": { "value": 3, "left": { "value": 1, "left": nil, "right": nil }, "right": { "value": 2, "left": { "value": -2, "left": nil, "right": nil }, "right": { "value": -3, "left": nil, "right": nil } } } }, s: 7, x: true },
{ t: { "value": 4, "left": { "value": 1, "left": { "value": -2, "left": nil, "right": { "value": 3, "left": nil, "right": nil } }, "right": nil }, "right": { "value": 3, "left": { "value": 1, "left": nil, "right": nil }, "right": { "value": 2, "left": { "value": -4, "left": nil, "right": nil }, "right": { "value": -3, "left": nil, "right": nil } } } }, s: 7, x: false },
{ t: { "value": 4, "left": { "value": 1, "left": { "value": -2, "left": nil, "right": { "value": 3, "left": nil, "right": nil } }, "right": nil }, "right": { "value": 3, "left": { "value": 1, "left": { value: -1, left: nil, right: nil}, "right": nil }, "right": { "value": 2, "left": { "value": -4, "left": nil, "right": nil }, "right": { "value": -3, "left": nil, "right": nil } } } }, s: 7, x: true },
{ t: nil, s: 0, x: true },
{ t: nil, s: 1, x: false },
{ t: { "value": 5, "left": nil, "right": nil }, s: 5, x: true },
{ t: { "value": 5, "left": nil, "right": nil }, s: -5, x: false },
{ t: { "value": 5, "left": nil, "right": nil }, s: 0, x: false },
{ t: { "value": 8, "left": nil, "right": { "value": 3, "left": nil, "right": nil } }, s: 8, x: false },
].each do |x|
it do
expect(path_with_given_sum?(Tree.build_from(x[:t]), x[:s])).to eql(x[:x])
end
end
class Tree
attr_accessor :value, :left, :right
def initialize(value, left: nil, right: nil)
@value = value
@left = left
@right = right
end
def to_h
{ value: value, left: left&.to_h, right: right&.to_h }
end
def self.build_from(hash)
return nil if hash.nil?
Tree.new(hash[:value], left: build_from(hash[:left]), right: build_from(hash[:right]))
end
end
end
|
