# Share my clear and concise JAVA DFS solution

• Note that if an 'O' is not surrounded by 'X', then this 'O' must at the edge of the board or has connected with an edge_board_O, so what we need to do is to find ALL 'O's in this kind and mark them. So I ues DFS to find them and mark them as 'M'.

In the DFS_Edge_O_Marking function, I put the if statement e.g. if(i-1>0) to reduce the recrusion times in case of stack over flow.

``````public class Solution {
public void solve(char[][] board) {
if(board==null) return;
int m=board.length;
if(m==0) return;
int n=board[0].length;
for(int i=0;i<m;i++)
for(int j=0;j<n;j++){
if(i==0||j==0||i==m-1||j==n-1){DFS_Edge_O_Marking(board,i,j);
}
}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++){
if(board[i][j]=='M') board[i][j]='O';
else board[i][j]='X';
}
return;

}

public void DFS_Edge_O_Marking(char[][] board,int i,int j) {
if(i<0||j<0||i>=board.length||j>=board[0].length) return;
if(board[i][j]=='M'||board[i][j]=='X') return;
board[i][j]='M';
if(i-1>0) DFS_Edge_O_Marking(board,i-1,j);//put a if statement in case of stack over flow;
if(i+1<board.length) DFS_Edge_O_Marking(board,i+1,j);
if(j-1>0) DFS_Edge_O_Marking(board,i,j-1);
if(j+1<board[0].length) DFS_Edge_O_Marking(board,i,j+1);
}
}
``````

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