<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]:
        dir = True
        r, c = 0, 0
        A, B = len(mat), len(mat[0])
        res = []
        for i in range(A * B):
            res.append(mat[r][c])
            if dir:
                if r == 0 and c != B - 1:
                    dir = False
                    c += 1
                elif c == B - 1:
                    dir = False
                    r += 1
                else:
                    r -= 1
                    c += 1
            else:
                if c == 0 and r != A - 1:
                    dir = True
                    r += 1
                elif r == A - 1:
                    dir = True
                    c += 1
                else:
                    c -= 1
                    r += 1
            
        return res
            
            
            
19800