# Why does DFS work in solving this problem?

• In the problem Surrounded Regions, DFS may yield stack overflow, why does it work in this problem? Anybody ever thought about it?

• If you keep track of what node has been visited, the maximum memory space will just be o(n).

``````void dfs(vector<vector<char>> &board, int a, int b){
stack<pair<int, int>>dp;
dp.push(make_pair(a, b));
while (!dp.empty()){
pair<int, int> temp = dp.top();
dp.pop();
int x = temp.first;
int y = temp.second;
if (x < 0 || y < 0 || x >= board.size() || y >= board.front().size())continue;
if (board[x][y] == 'O'){
board[x][y] = 'T';
}
else{
continue;
}
dp.push(make_pair(x - 1, y));
dp.push(make_pair(x + 1, y));
dp.push(make_pair(x, y - 1));
dp.push(make_pair(x, y + 1));
}
}
void solve(vector<vector<char>> &board) {
if (board.empty())return;
for (int i = 0; i < board.front().size(); ++i){
dfs(board, 0, i);
dfs(board, board.size() - 1,i );
}
for (int i = 0; i < board.size(); ++i){
dfs(board, i, 0);
dfs(board, i, board.front().size()-1);
}
for (int i = 0; i < board.size(); ++i){
for (int j = 0; j < board.front().size(); ++j){
if (board[i][j] == 'T'){
board[i][j] = 'O';
}
else{
board[i][j] = 'X';
}
}
}
}``````

• Do you mean using hash table or else to record each node visited? I fail to get your point.

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