边工作边刷题：70天一遍leetcode: day 90

Binary Tree Vertical Order Traversal

要点：这题不难，就是要找到最简单的方法还需要有些考量。主要的原因是Tree的宽度是不知道的

• 用map记录min->max的区间，最后再iterate build返回数组。这样可以不用多traverse一遍Tree。
• traversal可以是dfs或者bfs
• dfs/bfs时候要存结点的pos（有一种类似解法是创建一个2d map[col][height]来存结点）。这个的意思就是left/right结点的pos是依赖于上一层的，而不是global的

效率：

• keys: 24.83% => 42.28%，for k in cols vs for k in cols.keys()
• next vs. counting: 42.28% => 58.39%

https://repl.it/Ccf7/2

# 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, deque

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def verticalOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root: return []
        cols = defaultdict(list)
        q = deque([(root, 0)])
        while q:
            ql = len(q)
            for _ in xrange(ql):
                root, pos= q.popleft()
                cols[pos].append(root.val)
                if root.left:
                    q.append((root.left, pos-1))
                    
                if root.right:
                    q.append((root.right, pos+1))
        return [cols[k] for k in xrange(min(cols.keys()), max(cols.keys())+1)]