Another Casual Coder

Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation: 
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).

Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation: 
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6.

• First, a Sorted Dictionary sorted by the x-index
• Second, the value of the above data structure will be a Sorted List sorted by the y-index
• Because of this statement, the value of the above data structure still needs to be sorted, hence I'm using as the value a Sorted Set to allow for duplicates

public class Solution
{
	public IList<IList<int>> VerticalTraversal(TreeNode root)
	{
		SortedDictionary<int, SortedList<int, SortedSet<int>>> vertical = new SortedDictionary<int, SortedList<int, SortedSet<int>>>();
		VerticalTraversal(root, 0, 0, vertical);
		List<IList<int>> solution = new List<IList<int>>();
		foreach (int x in vertical.Keys)
		{
			List<int> list = new List<int>();
			foreach (int y in vertical[x].Keys)
				foreach (int element in vertical[x][y])
					list.Add(element);
			solution.Add(list);
		}
		return solution;
	}
	private void VerticalTraversal(TreeNode root,
									int x,
									int y,
									SortedDictionary<int, SortedList<int, SortedSet<int>>> vertical)
	{
		if (root == null) return;
		if (!vertical.ContainsKey(x)) vertical.Add(x, new SortedList<int, SortedSet<int>>());
		if (!vertical[x].ContainsKey(y)) vertical[x].Add(y, new SortedSet<int>());
		vertical[x][y].Add(root.val);
		VerticalTraversal(root.left, x - 1, y + 1, vertical);
		VerticalTraversal(root.right, x + 1, y + 1, vertical);
	}
}