<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>ans;

void f(vector<vector<int>>&v,int i,int j,int dir){
if(i>=v.size()||j>=v[0].size()){
    return;
}
ans.push_back(v[i][j]);
if(dir==1){
if(i-1>=0&&j+1<v[0].size()){
f(v,i-1,j+1,dir);
}
else{
if(j+1<v[0].size()){ f(v,i,j+1,2); return ;}
f(v,i+1,j,2);
}

}
else{

if(j-1>=0&&i+1<v.size()){
f(v,i+1,j-1,dir);
}
else{
if(i+1<v.size()){f(v,i+1,j,1);return;}
f(v,i,j+1,1);
}



}

return ;
}
    vector<int> findDiagonalOrder(vector<vector<int>>& v) {
        
f(v,0,0,1);
return ans;
    }
};
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