human
Solve in Python:
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.
 Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.

# 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]]:
        if root is None:
            return []

        columnTable = defaultdict(list)
        min_column = max_column = 0

        def BFS(root):
            nonlocal min_column, max_column
            queue = deque([(root, 0, 0)])

            while queue:
                node, row, column = queue.popleft()

                if node is not None:
                    columnTable[column].append((row, node.val))
                    min_column = min(min_column, column)
                    max_column = max(max_column, column)

                    queue.append((node.left, row + 1, column - 1))
                    queue.append((node.right, row + 1, column + 1))

        # step 1). BFS traversal
        BFS(root)
        
         # step 2). extract the values from the columnTable
        ret = []
        for col in range(min_column, max_column + 1):
            # sort first by 'row', then by 'value', in ascending order
            ret.append([val for row, val in sorted(columnTable[col])])

        return ret