【牛客刷题】bfs和dfs （二叉树层序遍历、矩阵最长递增路径、被围绕的区域）

二叉树层序遍历

   vector<vector<int> > levelOrder(TreeNode* root) {
        // write code here
        vector<int> res;
        vector<vector<int>> result;
        if (root == nullptr) return result;
        queue<TreeNode*> que;
        que.push(root);

        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* temp = que.front();
                que.pop();
                res.push_back(temp->val);
                if (temp->left) que.push(temp->left);
                if (temp->right) que.push(temp->right);
            }
            result.push_back(res);
            res.clear();
        }
        return result;
    }
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矩阵最长递增路径

https://www.nowcoder.com/share/jump/9321389651694076681305
BFS 通常是为了找到最短路径，求最长路径最好用DFS！

拓扑排序（增加inDegrees矩阵）+ BFS

    int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    vector<vector<int>> getInDegrees(vector<vector<int> >& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> inDegrees(m, vector<int> (n, 0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < 4; k++) {
                    int newX = i + dir[k][0];
                    int newY = j + dir[k][1];
                    if (newX < 0 || newX >= m || newY < 0 || newY >= n) continue;
                    if (grid[i][j] > grid[newX][newY]) {
                        inDegrees[i][j]++;
                    }
                }
            }
        }
        return inDegrees;
    }

    int solve(vector<vector<int> >& matrix) {
        // write code here
        vector<vector<int>> inDegrees = getInDegrees(matrix);
        int maxLen = 0;
        queue<pair<int, int>> que;

        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[0].size(); j++) {
                if (inDegrees[i][j] == 0) {
                    que.push({i, j});
                }
            }
        }

        while (!que.empty()) {
            maxLen++;
            int size = que.size();
            for (int i = 0; i < size; i++) {//需要处理每层信息时这样写，类似于二叉树的层序遍历
                int x = que.front().first;
                int y = que.front().second;
                que.pop();
                for (int k = 0; k < 4; k++) {//遍历方向
                    int newX = x + dir[k][0];
                    int newY = y + dir[k][1];
                    if (newX < 0 || newX >= matrix.size() || newY < 0 ||
                            newY >= matrix[0].size()) continue;
                    if (matrix[x][y] < matrix[newX][newY]) {//保证是递增序列
                        inDegrees[newX][newY]--;//因为已经确保递增了，所以减少newX和newY的一个入度
                        if (inDegrees[newX][newY] == 0) {//当入度全为0，表示条件全满足，所以可以入队
                            que.push({newX, newY});
                        }
                    }

                }
            }
        }
        return maxLen;
    }
};
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单纯的bfs：

    int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    int bfs(vector<vector<int>>& matrix, vector<vector<bool>>& visited, int x,
             int y) {
        queue<pair<int, int>> que;
        que.push({x, y});
        visited[x][y] = true;
        int maxLen = 0;

        while (!que.empty()) {
            maxLen++;
            int size = que.size();
            for (int i = 0; i < size; i++) {//层处理
                int curX = que.front().first;
                int curY = que.front().second;
                que.pop();
                for (int k = 0; k < 4; k++) {//方向处理
                    int newX = curX + dir[k][0];
                    int newY = curY + dir[k][1];
                    if (newX < 0 || newX >= matrix.size() || newY < 0 ||
                            newY >= matrix[0].size()) continue;
                    if (!visited[newX][newY] && matrix[curX][curY] < matrix[newX][newY]) {
                        que.push({newX, newY});
                        visited[newX][newY] = true;
                    }
                }
            }
        }
        return maxLen;
    }

    int solve(vector<vector<int> >& matrix) {
        // write code here
        vector<vector<bool>> visited(matrix.size(), vector<bool>(matrix[0].size(),
                                     false));
        int maxLen = 0;

        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[0].size(); j++) {
                maxLen = max(maxLen, bfs(matrix, visited, i, j));
            }
        }
        return maxLen;
    }
    
};
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dfs+记忆化搜索（memo）：

    int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
    int dfs(vector<vector<int>>& matrix, vector<vector<int>>& memo, int x, int y) {
        if (memo[x][y] != -1) return memo[x][y];//递归终止条件
        int maxLen = 1;

        for (int i = 0; i < 4; i++) {//遍历方向
            int newX = x + dir[i][0];
            int newY = y + dir[i][1];
            if (newX < 0 || newX >= matrix.size() || newY < 0 || newY >= matrix[0].size() ||
                    matrix[newX][newY] <= matrix[x][y]) continue;//满足条件才递归
            maxLen = max(maxLen, 1 + dfs(matrix, memo, newX, newY));//表示从(x, y)到(newX, newY)这一步
        }
        memo[x][y] = maxLen;
        return memo[x][y];
    }

    int solve(vector<vector<int> >& matrix) {
        vector<vector<int>> memo(matrix.size(), vector<int>(matrix[0].size(), -1));
        int maxLen = 0;

        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[0].size(); j++) {
                maxLen = max(maxLen, dfs(matrix, memo, i, j));
            }
        }
        return maxLen;
    }
};
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被围绕的区域

https://www.nowcoder.com/share/jump/9321389651694087623428

dfs:

    int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
    void dfs(vector<vector<char>>& matrix, int x,
             int y) {
        int m = matrix.size(), n = matrix[0].size();
        matrix[x][y] = 'E';

        for (int i = 0; i < 4; i++) {
            int newX = x + dir[i][0];
            int newY = y + dir[i][1];
            if (newX < 0 || newX >= m || newY < 0 || newY >= n ||
                    matrix[newX][newY] != 'O')
                continue;//(newX,newY)中超出边界的、不是O的不用管
            dfs(matrix, newX, newY);
        }
    }

    vector<vector<char> > surroundedArea(vector<vector<char> >& board) {
        // write code here
        int m = board.size(), n = board[0].size();
        //将连接边界的O全部替换
        for (int i = 0; i < m; i++) {
            if (board[i][0] == 'O') dfs(board, i, 0);
            if (board[i][n - 1] == 'O') dfs(board, i, n - 1);
        }
        for (int j = 0; j < n; j++) {
            if (board[0][j] == 'O') dfs(board, 0, j);
            if (board[m - 1][j] == 'O') dfs(board, m - 1, j);
        }
        //又替换回来
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'E') board[i][j] = 'O';
                else board[i][j] = 'X';
            }
        }
        return board;
    }
};
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bfs:

    int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};

    void bfs(vector<vector<char>>& matrix, int x, int y) {
        int m = matrix.size(), n = matrix[0].size();
        queue<pair<int, int>> que;
        que.push({x, y});
        while (!que.empty()) {
            int curX = que.front().first;
            int curY = que.front().second;
            que.pop();
            matrix[curX][curY] = 'E';
            for (int i = 0; i < 4; i++) {
                int newX = curX + dir[i][0];
                int newY = curY + dir[i][1];
                if (newX < 0 || newX >= m || newY < 0 || newY >= n ||
                        matrix[newX][newY] != 'O')
                    continue;//(newX,newY)中超出边界的、不是O的不用管
                que.push({newX, newY});
            }
        }
    }

    vector<vector<char> > surroundedArea(vector<vector<char> >& board) {
        // write code here
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; i++) {
            if (board[i][0] == 'O') bfs(board, i, 0);
            if (board[i][n - 1] == 'O') bfs(board, i, n - 1);
        }
        for (int j = 0; j < n; j++) {
            if (board[0][j] == 'O') bfs(board, 0, j);
            if (board[m - 1][j] == 'O') bfs(board, m - 1, j);
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'E') board[i][j] = 'O';
                else board[i][j] = 'X';
            }
        }
        return board;
    }
};
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