Given a binary tree, return the vertical order traversal of its nodes values.
For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).
Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).
If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.
Return an list of non-empty reports in order of X coordinate.  Every report will have a list of values of nodes.
Example 1:
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).
Example 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.
Note:
The tree will have between 1 and 1000 nodes.
Each node's value will be between 0 and 1000.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
        hmap = defaultdict(list)
        x_min, x_max = 0, 0
        queue = deque()
        report = list()
        if root:
            queue.append((root, 0, 1))
        else:
            return report
        
        while queue:
            root, x, y = queue.popleft()
            
            x_min = min(x, x_min)
            x_max = max(x, x_max)
            while y > len(hmap[x]):
                hmap[x].append([])
            hmap[x][-1].append(root.val)
                
            if root.left:
                queue.append((root.left, x - 1, y + 1))
            if root.right:
                queue.append((root.right, x + 1, y + 1))
                
        print(hmap) 
        print(x_min, x_max)
        for i in range(x_min, x_max + 1):
            report.append([])
            for l in hmap[i]:
                print(l)
                print(report)
                if l:
                    l.sort()
                    report[-1].extend(l)
        return report
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
        hmap = defaultdict(list)
        x_min, x_max = 0, 0
        queue = deque()
        report = list()
        if root:
            queue.append((root, 0, 1))
        else:
            return report
        
        while queue:
            root, x, y = queue.popleft()
            
            x_min = min(x, x_min)
            x_max = max(x, x_max)
            while y > len(hmap[x]):
                hmap[x].append([])
            hmap[x][-1].append(root.val)
                
            if root.left:
                queue.append((root.left, x - 1, y + 1))
            if root.right:
                queue.append((root.right, x + 1, y + 1))
                
        print(hmap) 
        print(x_min, x_max)
        for i in range(x_min, x_max + 1):
            report.append([])
            for l in hmap[i]:
                print(l)
                print(report)
                if l:
                    l.sort()
                    report[-1].extend(l)
        return report