<p>Given a 2D integer array <code>nums</code>, return <em>all elements of </em><code>nums</code><em> in diagonal order as shown in the below images</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_1_1784.png" style="width: 158px; height: 143px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,4,2,7,5,3,8,6,9]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/08/sample_2_1784.png" style="width: 230px; height: 177px;" />
<pre>
<strong>Input:</strong> nums = [[1,2,3,4,5],[6,7],[8],[9,10,11],[12,13,14,15,16]]
<strong>Output:</strong> [1,6,2,8,7,3,9,4,12,10,5,13,11,14,15,16]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i].length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= sum(nums[i].length) &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& nums) {
        int n = nums.size();
         int m = -1;

        vector<vector<bool>> visited;

        for(int i=0;i<n;i++){
            int m = nums[i].size();
            vector<bool> temp(m,false);
            visited.push_back(temp);
        }

        visited[0][0] = true;

        queue<pair<int,int>> q;

        q.push({0,0});

        int drow[] = {1,0};
        int dcol[] = {0,1};

        vector<int> output;

        while(!q.empty()){
          
          auto it = q.front();

          q.pop();

          int row = it.first;
          int col = it.second;

          output.push_back(nums[row][col]);

          for(int i=0;i<2;i++){
            int nrow = row + drow[i];
            int ncol = col + dcol[i];

            if(nrow>=0 && nrow<n && ncol>=0 && ncol<nums[nrow].size()  && visited[nrow][ncol]==false){
                visited[nrow][ncol] = true;
                q.push({nrow,ncol});
            }
          }
        }

        return output;
    }
};
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