<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();
        int n = mat[0].size();
        vector<int> result;
        result.reserve(m * n);  // Pre-allocate space
        
        for (int sum = 0; sum < m + n - 1; sum++) {
            int row = sum < n ? 0 : sum - n + 1;
            int col = sum < n ? sum : n - 1;
            
            vector<int> diagonal;
            diagonal.reserve(min(m, n));  // Pre-allocate space
            
            while (row < m && col >= 0) {
                diagonal.push_back(mat[row][col]);
                row++;
                col--;
            }
            
            if (sum % 2 == 0) {
                reverse(diagonal.begin(), diagonal.end());
            }
            
            result.insert(result.end(), diagonal.begin(), diagonal.end());
        }
        
        return result;
    }
};
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