<p>Given an <code>m x n</code> matrix <code>mat</code>, return <em>an array of all the elements of the array in a diagonal order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/10/diag1-grid.jpg" style="width: 334px; height: 334px;" />
<pre>
<strong>Input:</strong> mat = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,4,7,5,3,6,8,9]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,2],[3,4]]
<strong>Output:</strong> [1,2,3,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def findDiagonalOrder(self, mat: List[List[int]]) -> List[int]:
        matrix = mat
        if not matrix or not matrix[0]:
            return []

        def get_all_on_diagonal(matrix, diagonal_index):
            # Return elements on diagonal
            elements = []
            #diag = row + col
            for row in range(len(matrix)):
                col = diagonal_index - row
                if 0 <= col < len(matrix[row]):
                    elements.append(matrix[row][col])
                
            return elements

        m = len(matrix)
        n = len(matrix[0])
        result = []

        for diagonal_index in range(m+n-1):
            elements = get_all_on_diagonal(matrix, diagonal_index)

            # Alternate every other one 
            if diagonal_index % 2 != 1:
                elements.reverse()
            result.extend(elements)
        return result

        

        
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