Problem description:

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 a list of non-empty reports in order of X coordinate. Every report will have a list of values of nodes.

Example 1:

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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).


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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.

Solution:

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# 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]]:
        res = collections.defaultdict(list)
        queue = [(root, 0)]
        while queue:
            nextLevelQueue = []
            curLevelRes = collections.defaultdict(list)
            for node, level in queue:
                curLevelRes[level].append(node.val)
                if node.left: 
                    nextLevelQueue += (node.left, level-1),
                if node.right: 
                    nextLevelQueue += (node.right, level+1),
            for level in curLevelRes: res[level].extend(sorted(curLevelRes[level]))
            queue = nextLevelQueue
        return [res[level] for level in sorted(res)]

time complexity: $O()$
space complexity: $O()$
reference:
related problem: