<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:
    using Pair = pair<int,int>;
    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        vector<int> res;
        bool dir = true;
        vector<Pair> cur = {{0,0}};
        while(cur.size()) {
            vector<Pair> next;
            for (int i=0; i<cur.size(); i++) {
                int x = cur[i].first;
                int y = cur[i].second;
                if (i==0) {
                    if (x<m-1) {
                        next.push_back({x+1,y});
                    }
                }
                if (y < n-1) {
                    next.push_back({x,y+1});
                }
            }

            if (dir) {
                for (int i=0; i<cur.size(); i++) {
                    res.push_back(mat[cur[i].first][cur[i].second]);
                }
            } else {
                for (int i=cur.size()-1; i>=0; i--) {
                    res.push_back(mat[cur[i].first][cur[i].second]);
                }
            }

            cur = next;
            dir = !dir;
        }

        return res;
    }
};
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