<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]:
        if not mat or not mat[0]:
            return []
    
        m, n = len(mat), len(mat[0])
        res = []
        temp = []

        # Iterate over all possible values of the sum of indices (diagonal levels)
        for s in range(m + n - 1):
            temp.clear()  # Clear the temporary list for the next diagonal

            # Determine the starting point of the diagonal
            start_row = 0 if s < n else s - n + 1
            start_col = s if s < n else n - 1

            # Collect elements of the current diagonal
            while start_row < m and start_col >= 0:
                temp.append(mat[start_row][start_col])
                start_row += 1
                start_col -= 1
            
            # Append to result based on even or odd diagonal level
            if s % 2 == 0:
                res.extend(reversed(temp))
            else:
                res.extend(temp)

        return res
19500