<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:
    def findDiagonalOrder(self, nums: List[List[int]]) -> List[int]:
        """
            Time: O(N x M)
        """
        queue = deque([(0,0)])
        result = []
        visited = {}
        visited[(0,0)] = True
        while len(queue):
            currRow, currCol = queue.popleft()
            result.append(nums[currRow][currCol])
            for edgeRow, edgeCol in [(1,0), (0,1)]:
                newCoord = (currRow + edgeRow, currCol + edgeCol)
                if newCoord in visited:
                    continue
                if newCoord[0] <= len(nums) - 1 and newCoord[1] <= len(nums[newCoord[0]]) - 1:
                    visited[newCoord] = True
                    queue.append(newCoord)

        return result

        
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