<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 n = mat.size();
        int m = mat[0].size();
        int topDown = 0;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        queue<pair<pair<int, int>, int>> q;
        q.push({{0, 0}, topDown});
        vis[0][0] = 1;
        vector<pair<int, int>> ans;

        while (!q.empty()) {
            int row = q.front().first.first;
            int col = q.front().first.second;
            int flag = q.front().second;
            q.pop();
            
            ans.push_back({mat[row][col], flag});
            
            if (col + 1 < m && !vis[row][col + 1]) {
                q.push({{row, col + 1}, !flag});
                vis[row][col + 1] = 1;
            }

            if (row + 1 < n && !vis[row + 1][col]) {
                q.push({{row + 1, col}, !flag});
                vis[row + 1][col] = 1;
            }
        }
        
        vector<int> res;
        int i = 0;

        while (i < ans.size()) {
            if (ans[i].second == 1) {
                // Collect all elements in this diagonal in top-down order
                while (i < ans.size() && ans[i].second == 1) {
                    res.push_back(ans[i].first);
                    i++;
                }
            } else {
                // Collect all elements in this diagonal in bottom-up order
                int l = i;
                while (i < ans.size() && ans[i].second == 0) {
                    res.push_back(ans[i].first);
                    i++;
                }
                reverse(res.begin()+l,res.begin()+i);
            }
        }

        return res;
    }
};

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