# 3 solutions: iterative version of BFS and DFS, recursive version of DFS, C++ implementation

• BFS iterative solution using Queue:

``````struct GridPoint {
int x;
int y;
GridPoint(int _x=0, int _y=0) : x(_x), y(_y) {}
};

class Solution {
public:
int numIslands(vector<vector<char> > & grid) {
if (grid.size() == 0) return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int> > visited(m, vector<int>(n, 0));
return bfs(grid, visited, m, n);
}

int bfs(vector<vector<char> > & grid, vector<vector<int> > & visited, int m, int n) {
int numLands = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
visited[i][j] = 1;
queue<GridPoint> q;
q.push(GridPoint(i, j));
while (!q.empty()) {
GridPoint tmp = q.front(); q.pop();
//if right is feasible
if (tmp.y + 1 < n && !visited[tmp.x][tmp.y+1]) {
visited[tmp.x][tmp.y + 1] = 1;
if (grid[tmp.x][tmp.y + 1] == '1')
q.push(GridPoint(tmp.x, tmp.y + 1));
}
//if down is feasible
if (tmp.x + 1 < m && !visited[tmp.x + 1][tmp.y]) {
visited[tmp.x + 1][tmp.y] = 1;
if (grid[tmp.x + 1][tmp.y] == '1')
q.push(GridPoint(tmp.x + 1, tmp.y));
}
//pay attention: left should also be considered!
if (tmp.y - 1 >= 0 && !visited[tmp.x][tmp.y - 1]) {
visited[tmp.x][tmp.y - 1] = 1;
if (grid[tmp.x][tmp.y - 1] == '1')
q.push(GridPoint(tmp.x, tmp.y-1));
}
//pay attention: up should also be considered!
if (tmp.x - 1 >= 0 && !visited[tmp.x - 1][tmp.y]) {
visited[tmp.x - 1][tmp.y] = 1;
if(grid[tmp.x - 1][tmp.y] == '1')
q.push(GridPoint(tmp.x - 1, tmp.y));
}
}
numLands++;
}
}
}

return numLands;
}
};
``````

DFS iterative solution using stack:

``````struct GridNode {
int x;
int y;
GridNode(int _x=0, int _y=0) : x(_x), y(_y) {}
};

//dfs iterative solution
class Solution {
public:
int numIslands(vector<vector<char> > & grid) {
if (grid.size() == 0) return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int> > visited(m, vector<int>(n, 0));
return dfs(grid, visited, m, n);
}

int dfs(vector<vector<char> > & grid, vector<vector<int> > & visited, int m, int n) {
int numLands = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
visited[i][j] = 1;
stack<GridNode> s;
s.push(GridNode(i, j));

while (!s.empty()) {
GridNode curNode = s.top();
GridNode nextNode;
bool canContinue = true;
//check if current node can continue to go to next
//if can, choose one adjacent node to push to stack
if (curNode.x + 1 < m && !visited[curNode.x + 1][curNode.y] && grid[curNode.x + 1][curNode.y] == '1') {
nextNode = GridNode(curNode.x + 1, curNode.y);
} else if (curNode.y + 1 < n && !visited[curNode.x][curNode.y + 1] && grid[curNode.x][curNode.y + 1] == '1') {
nextNode = GridNode(curNode.x, curNode.y + 1);
} else if (curNode.x - 1 >= 0 && !visited[curNode.x - 1][curNode.y] && grid[curNode.x - 1][curNode.y] == '1') {
nextNode = GridNode(curNode.x - 1, curNode.y);
} else if (curNode.y - 1 >= 0 && !visited[curNode.x][curNode.y - 1] && grid[curNode.x][curNode.y - 1] == '1') {
nextNode = GridNode(curNode.x, curNode.y - 1);
} else {
canContinue = false;
}

if(canContinue) {
visited[nextNode.x][nextNode.y] = 1;
s.push(nextNode);
curNode = nextNode;
continue;
}

//if not, pop the top element of the stack
s.pop();

}

numLands++;
}
}
}

return numLands;
}
};
``````

DFS recursive version:

``````class Solution {
public:
int numIslands(vector<vector<char> >& grid) {
if (grid.size() == 0) return 0;
int m = grid.size();
int n = grid[0].size();
int numLands = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == '1') {
dfs(grid, i, j, m, n);
numLands++;
}
}
}
return numLands;
}

void dfs(vector<vector<char> >& grid, int x, int y, int m, int n) {
grid[x][y] = '2'; //2 indicates that this position is visited
//visit all node in adjacent list
if (x + 1 < m && grid[x + 1][y] == '1') {
dfs(grid, x + 1, y, m, n);
}

if (y + 1 < n && grid[x][y + 1] == '1') {
dfs(grid, x, y + 1, m, n);
}

if (x - 1 >= 0 && grid[x - 1][y] == '1') {
dfs(grid, x-1, y, m, n);
}

if (y - 1 >= 0 && grid[x][y - 1] == '1') {
dfs(grid, x, y-1, m, n);
}
}
};
``````

Actually BFS cannot be implemented using recursion, if anyone knows how, please tell me? Thanks!

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