<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) {
        int m = mat.size(), n = mat[0].size();
        bool to_reverse = false;
        for (int i = 0; i < m; i++) {
            vector<int> line_result;
            int x = i, y = 0;
            get_line_result(mat, i, 0, to_reverse);
        }
        for (int i = 1; i < n; i++) {
            // I only need to give the start point
            get_line_result(mat, m-1, i, to_reverse);            
        }
        return result;
    }

    void get_line_result(vector<vector<int>>& mat, int x, int y, bool & to_reverse) {
        vector<int> line_result;
        int m = mat.size(), n = mat[0].size();

        while (x >= 0 && y < n) {
            line_result.push_back(mat[x][y]);
            x--; y++;
        }
        if (to_reverse) {
            reverse(line_result.begin(), line_result.end());
        }
        for (int v : line_result) {
            result.push_back(v);
        }
        to_reverse = !to_reverse;
    }

    vector<int> result;
};
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