Diagonal Traverse II

class Solution {
public:
    vector<int> traverseDiagonally(vector<vector<int>>& matrix) {
        queue<pair<int, int>> traversalQueue;
        traversalQueue.push({0, 0});
        vector<int> result;
        
        while (!traversalQueue.empty()) {
            auto [r, c] = traversalQueue.front();
            traversalQueue.pop();
            result.push_back(matrix[r][c]);
            
            if (c == 0 && r + 1 < matrix.size()) {
                traversalQueue.push({r + 1, c});
            }
            
            if (c + 1 < matrix[r].size()) {
                traversalQueue.push({r, c + 1});
            }
        }
        
        return result;
    }
};

In this solution for the "Diagonal Traverse II" problem using C++, the approach involves using a queue to manage the traversal of a given matrix diagonally. Here's a breakdown of how the implementation works:

• A queue is initialized to keep track of matrix positions during traversal. The top-left position (0, 0) is added initially to start the process.
• An empty vector named result is prepared to store the values as they are encountered.
• The queue is processed in a loop until it is empty. In each iteration:
  • The first position in the queue is dequeued to get the current row (r) and column (c).
  • The value at this position in the matrix is added to the result vector.
  • The logic then considers the next diagonal steps by evaluating the boundaries of the matrix. It checks for movement in two possible directions:
    1. Downward (next row, same column): Push the position (r + 1, c) to the queue if the next row exists within the matrix bounds.
    2. Rightward (same row, next column): Push the position (r, c + 1) to the queue if the next column exists within the current row boundaries.
• This loop continues until all possible paths in the matrix are explored and all the positions on the diagonals visited are pushed to the result vector.
• Finally, the vector result is returned, containing the values of the matrix as traversed diagonally according to the devised queuing mechanism.

This approach effectively captures diagonal traversal in a structured matrix, ensuring every potential “next step” is explored by adhering to the matrix’s boundary constraints.

class Solution {
    public int[] diagonalTraversal(List<List<Integer>> matrix) {
        Queue<Pair<Integer, Integer>> processingQueue = new LinkedList();
        processingQueue.offer(new Pair(0, 0));
        List<Integer> resultList = new ArrayList();
        
        while (!processingQueue.isEmpty()) {
            Pair<Integer, Integer> current = processingQueue.poll();
            int currentRow = current.getKey();
            int currentCol = current.getValue();
            resultList.add(matrix.get(currentRow).get(currentCol));
            
            if (currentCol == 0 && currentRow + 1 < matrix.size()) {
                processingQueue.offer(new Pair(currentRow + 1, currentCol));
            }
            
            if (currentCol + 1 < matrix.get(currentRow).size()) {
                processingQueue.offer(new Pair(currentRow, currentCol + 1));
            }
        }
        
        int[] diagonalOrder = new int[resultList.size()];
        int idx = 0;
        for (int value : resultList) {
            diagonalOrder[idx] = value;
            idx++;
        }
        
        return diagonalOrder;
    }
}

The given Java code solves the problem of traversing a list of lists (representing a matrix) diagonally. The solution uses a breadth-first search (BFS) approach embodied by a queue to manage how each element is visited. Below is a breakdown of the solution workflow:

• Initialize a queue to manage pairs of indexes representing positions in the matrix.
• Start with the top-left corner of the matrix, the position (0,0).
• Utilize a while loop to process each element until the queue is empty. In each iteration:
  • Dequeue a pair to determine the current position.
  • Store the value at the current position in the resultList.
  • If feasible (without going out of bounds), enqueue the position directly below (next row, same column).
  • Similarly, enqueue the position to the right (same row, next column), provided it is valid.
• After processing all elements, convert the resultList into an array for the final output.

This method ensures that the matrix is traversed in a zigzag or diagonal pattern, consequently adding those elements first whose row and column indices sum is smaller, simulating a diagonal traversal from the top left to the bottom right of the matrix. The use of a queue ensures each cell is processed in the correct sequence without backtracking.

class Solution:
    def diagonalTraversal(self, matrix: List[List[int]]) -> List[int]:
        q = deque([(0, 0)])
        result = []
        
        while q:
            r, c = q.popleft()
            result.append(matrix[r][c])
            
            if c == 0 and r + 1 < len(matrix):
                q.append((r + 1, c))
                
            if c + 1 < len(matrix[r]):
                q.append((r, c + 1))
        
        return result

In this Python3 solution, the function diagonalTraversal navigates a two-dimensional matrix diagonally, collecting values in a sequence. Ensure you provide the correct matrix input, which should be a list of lists of integers. This traversal strategy is achieved using a deque for efficient element insertion and removal.

Here's a breakdown of the code functionality:

• Initialize a deque q starting with the top-left corner of the matrix (0, 0).
• Create an empty list result that will store integers in the sequence of diagonal traversal.
• Use a while loop that continues as long as there are elements in q.
  • Remove the element at the front of the deque (popleft) which gives the current row and column indices, r and c.
  • Append the current matrix value matrix[r][c] to result.
  • If the current column is zero and there's a row below, append that next row's initial element to q.
  • If there's another column in the current row, append that next column's element to q.

The function ultimately returns the result list, which includes the matrix elements in the order of their diagonal traversal. This approach is robust and handles different matrix shapes, ensuring diagonal elements are processed correctly.