Breadth First Search: Shortest Reach (BFS) - ios

Question

https://www.hackerrank.com/challenges/bfsshortreach
My strategy is to keep an array of the Node and its corresponding adjacent nodes using an integer and unordered set.
Then I start at the starting node and iterate throughout, keeping a set of all the corresponding adjacent nodes and terminating if I visit any node I've already visited.
Thanks for any Swifters willing to help me! Spent a day trying to figure it out and it's driving me crazy. :(
let testCases : Int = Int(readLine()!)!
for i in 0..<testCases
{
let inputArray : [Int] = String(readLine()!).characters.split(" ").map{Int(String($0))!}
let numberOfNodes : Int = inputArray[0]
let numberOfEdges : Int = inputArray[1]
var dictionary : [Int : Set<Int>] = [:]
var countDictionary : [Int : Int] = [:]
for j in 0..<numberOfEdges
{
let edge : [Int] = String(readLine()!).characters.split(" ").map{Int(String($0))!}
let start : Int = edge[0]
let end : Int = edge[1]
if let val = dictionary[start]
{
dictionary[start]!.insert(end)
}
else
{
dictionary[start] = Set.init(arrayLiteral: end)
}
if let val = dictionary[end]
{
dictionary[end]!.insert(start)
}
else
{
dictionary[end] = Set.init(arrayLiteral: start)
}
}
let temp : [Int] = String(readLine()!).characters.split(" ").map{Int(String($0))!}
let startPoint : Int = temp[0]
var traversed : Set<Int> = Set()
func iterate(node : Int, _ count : Int)
{
let newCount = count + 6
var temp : [Int] = []
if let val = dictionary[node]
{
for i in val
{
temp.append(i)
}
}
if let val = countDictionary[node]
{
if count < val
{
countDictionary[node]! = count
}
}
else
{
countDictionary[node] = count
}
if traversed.contains(node)
{
return
}
traversed.insert(node)
for j in temp
{
iterate(j, newCount)
}
}
iterate(startPoint, 0)
for z in 0..<numberOfNodes
{
if let val = countDictionary[z + 1]
{
if val > 0
{
print(val, terminator: " ")
}
}
else
{
print(-1, terminator: " ")
}
}
print(" ")
}