Leetcode 314. Binary Tree Vertical Order Traversal

Seawolf

Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column).

If two nodes are in the same row and column, the order should be from left to right.

Examples 1:

Input: [3,9,20,null,null,15,7]

   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7 

Output:

[
  [9],
  [3,15],
  [20],
  [7]
]
Examples 2:

Input: [3,9,8,4,0,1,7]

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7 

Output:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]
Examples 3:

Input: [3,9,8,4,0,1,7,null,null,null,2,5] (0's right child is 2 and 1's left child is 5)

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7
    /\
   /  \
   5   2

Output:

[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]

# 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 verticalOrder(self, root: TreeNode) -> List[List[int]]:
        res = []
        if root == None:
            return res
        hashmap = collections.defaultdict(list)
        queue = []
        cols = []
        max_val = 0
        min_val = 0
        
        queue.append(root)
        cols.append(0)
        while queue:
            curt = queue.pop(0)
            col = cols.pop(0)
            hashmap[col].append(curt.val)
            if curt.left != None:
                min_val = min(min_val, col - 1)
                queue.append(curt.left)
                cols.append(col - 1)
            if curt.right != None:
                max_val = max(max_val, col + 1)
                queue.append(curt.right)
                cols.append(col + 1)
        
        for i in range(min_val, max_val + 1):
            res.append(hashmap[i])
        return res