<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:
    void dfs(vector<vector<int>>& mat, vector<int>& sub, int rows, int cols, int r, int c) {
        if (r < 0 || c < 0 || r  > rows || c > cols) {
            return;
        }

        sub.push_back(mat[r][c]);
        dfs(mat, sub, rows, cols, r-1, c+1);
        return;
    }

    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        vector<int> ans;
        vector<vector<int>> storage;
        int left = 0, cols = mat[0].size()-1, top = 0, rows = mat.size()-1;
        
        // From top to bottom
        vector<int> sub;
        for (int i = top; i <= rows; i++) {
            sub.clear();
            dfs(mat, sub, rows, cols, i, 0);
            storage.push_back(sub);
        }
        // From left to right
        for (int i = left+1; i <= cols; i++) {
            sub.clear();
            dfs(mat, sub, rows, cols, rows, i);
            storage.push_back(sub);
        }


        for (int i = 0; i < storage.size(); i++) {
            if (i % 2 == 0) {
                // first to last
                for (int j = 0; j < storage[i].size(); j++) {
                    ans.push_back(storage[i][j]);
                }
            }
            else {
                // last to first
                for (int j = storage[i].size()-1; j >= 0; j--) {
                    ans.push_back(storage[i][j]);
                }
            }
        }

        return ans;
    }
};
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