LeetCode "Word Ladder II"

Man it took me numerous submissions to get AC, and I referred to this link: http://yihuad.blogspot.com/2013/11/word-ladder-ii-leetcode.html
Idea is pretty simple: since there's strict memory limit, we can simply organize the original dictionary as a DAG, by using BFS - by so, there could be only ONE copy of the whole dictionary. Then DFS could be used to retrieve all pathes.

class Solution {
public:
void buildPath(string curr, string end, vector<string> rec, vector<vector<string>> &ret, unordered_map<string, vector<string>> &dag)
{
rec.push_back(curr);
if (curr == end)
{
size_t currLen = rec.size();
if (ret.empty()) ret.push_back(rec);
else
{
size_t currRetLen = ret[0].size();
if (currLen < currRetLen)
{
ret.clear();
ret.push_back(rec);
}
else if (currLen == currRetLen)
{
ret.push_back(rec);
}
}
return;
}

vector<string> &nexts = dag[curr];
for (auto it = nexts.begin(); it != nexts.end(); it++)
{
vector<string> currrec = rec;
buildPath(*it, end, currrec, ret, dag);
}
}

vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
unordered_map<string, vector<string>> dag;
dict.insert(start); dict.insert(end);

vector<unordered_set<string>> q(2);
int cur = 0, next = 1;
q[cur].insert(start);
while (!q[cur].empty())
{
for (auto it = q[cur].begin(); it != q[cur].end(); it++)
dict.erase(*it);

for (auto it = q[cur].begin(); it != q[cur].end(); it++)
{
string str = *it;
for (int i = 0; i < str.length(); i++)
{
char tmp = str[i];
for (char k = 'a'; k <= 'z'; ++k)
{
if (k == tmp) continue;
str[i] = k;
if (dict.find(str) != dict.end())
{
dag[*it].push_back(str);
q[next].insert(str);
}
}
str[i] = tmp;
}
}

//    rolling array
q[cur].clear();
cur = cur == 0 ? 1 : 0;
next = next == 0 ? 1 : 0;
}

vector<vector<string>> ret;
vector<string> rec;
buildPath(start, end, rec, ret, dag);
return ret;
}
};