<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) {
        // 用款搜的手法强行level traverse
        // 其实每层reverse一下就行，但这里加工了一下，去掉了reverse的多余步骤。
        queue<pair<int, int>> q;
        int m = mat.size();
        int n = mat[0].size();
        vector<int> result(m * n, 0);
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        q.emplace(0, 0);
        visited[0][0] = true;
        int level = 0;
        int start = 0;;
        while (!q.empty()) {
            level++;
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto [x, y] = q.front();
                q.pop();
                if (level % 2 != 0) {
                    result[start + i] = mat[x][y];
                } else {
                    result[start + size - i - 1] = mat[x][y];
                }

                for (int j = 0; j < 2; j++) {
                    int a = x + dx[j];
                    int b = y + dy[j];
                    if (a < 0 || a >= m || b < 0 || b >= n || visited[a][b]) {
                        continue;
                    }
                    visited[a][b] = true;
                    q.emplace(a, b);
                }
            }
            start += size;
        }

        return result;
    }
private:
    int dx[2] = {1, 0};
    int dy[2] = {0, 1};
};
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