<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 {
public:

    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        if (mat.empty()) return {};
        int m = mat.size();
        int n = mat[0].size();
        vector<int> res;
        int x = 0, y = 0;
        
        int dir = 0;
        while (y < n) {
            vector<int> buf;
            int i = x, j = y;
            while (i >= 0 && j < n) {
                buf.push_back(mat[i][j]);
                i--;
                j++;
            }
            if (dir%2 == 0) {
                res.insert(res.end(), buf.begin(), buf.end());
            } else {
                res.insert(res.end(), buf.rbegin(), buf.rend());
            }
            dir++;

            if (x == m-1) {
                y++;
            } else {
                x++;
            }
        }
        
        return res;
    }
};
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