"""
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:

Given binary tree [3,9,20,null,null,15,7],
   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7
return its vertical order traversal as:
[
  [9],
  [3,15],
  [20],
  [7]
]
Given binary tree [3,9,8,4,0,1,7],
     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7
return its vertical order traversal as:
[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]
Given binary tree [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
return its vertical order traversal as:
[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]
"""


from collections import defaultdict
from collections import deque

from implementations.tree import BinaryTreeNode


def vertical_traversal(root):
    cache = defaultdict(list)
    queue = deque([(0, root)])
    min_level = 0
    while len(queue) > 0:
        level, curr = queue.popleft()
        min_level = min(min_level, level)
        cache[level].append(curr.value)
        if curr.left is not None:
            queue.append((level - 1, curr.left))
        if curr.right is not None:
            queue.append((level + 1, curr.right))

    result = []
    while min_level in cache:
        result.append(cache[min_level])
        min_level += 1
    return result


if __name__ == '__main__':
    tree = BinaryTreeNode(3)
    tree.left = BinaryTreeNode(9)
    tree.right = BinaryTreeNode(20)
    tree.right.left = BinaryTreeNode(15)
    tree.right.right = BinaryTreeNode(7)
    print vertical_traversal(tree)
