Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.

Example:

Input:
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
Output:  [1,2,4,7,5,3,6,8,9]
Explanation:

Note:

1. The total number of elements of the given matrix will not exceed 10,000.

I don't think this is a hard problem. It is easy to figure out the walk pattern. Anyway...
Walk patterns:

• If out of bottom border (row >= m) then row = m - 1; col += 2; change walk direction.
• if out of right border (col >= n) then col = n - 1; row += 2; change walk direction.
• if out of top border (row < 0) then row = 0; change walk direction.
• if out of left border (col < 0) then col = 0; change walk direction.
• Otherwise, just go along with the current direction.

Time complexity: O(m * n), m = number of rows, n = number of columns.
Space complexity: O(1).

public class Solution {
    public int[] findDiagonalOrder(int[][] matrix) {
        if (matrix == null || matrix.length == 0) return new int[0];
        int m = matrix.length, n = matrix[0].length;
        
        int[] result = new int[m * n];
        int row = 0, col = 0, d = 0;
        int[][] dirs = {{-1, 1}, {1, -1}};
        
        for (int i = 0; i < m * n; i++) {
            result[i] = matrix[row][col];
            row += dirs[d][0];
            col += dirs[d][1];
            
            if (row >= m) { row = m - 1; col += 2; d = 1 - d;}
            if (col >= n) { col = n - 1; row += 2; d = 1 - d;}
            if (row < 0)  { row = 0; d = 1 - d;}
            if (col < 0)  { col = 0; d = 1 - d;}
        }
        
        return result;
    }
}