<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 solve(vector<vector<int>>& mat,bool up,int i,int j){
    int n=mat.size();
    int m=mat[0].size();
    if(i==n-1 && j==m-1){
        ans.push_back(mat[i][j]);
        return;
    }

    ans.push_back(mat[i][j]);
    // cout<<n<<" "<<i<<" "<<m<<" "<<j<<endl;

    if(up){
        if(i-1<0 && j+1>=m){
           solve(mat,!up,i+1,j);
        }
        else if(i-1<0){
             solve(mat,!up,i,j+1);
        }
        else if(j+1>=m){
            solve(mat,!up,i+1,j);
        }
        else{
            solve(mat,up,i-1,j+1);
        }
    }
    else{
        if(j-1<0 && i+1>=n){
            solve(mat,!up,i,j+1);
        }
        else if(j-1<0){
            solve(mat,!up,i+1,j);
        }
        else if(i+1>=n){
            solve(mat,!up,i,j+1);
        }
        else{
            solve(mat,up,i+1,j-1);
        }
    }

    return;
}
    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        bool up=true;
       solve(mat,up,0,0);
       return ans;
    }
};
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