LeetCode #Lint651 2018-09-04

651. Binary Tree Vertical Order Traversal

https://www.lintcode.com/problem/binary-tree-vertical-order-traversal/description

这道题本质上还是一个层序遍历，只不过增加了一个纵向列的信息。每当有左孩子，添加当前列-1的列信息；每当有右孩子，添加当前列+1的列信息。最后将字典中，从min到max的所有列按顺序取出来即可。
代码如下：

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class TreeColumnNode:
    def __init__(self, node, col):
        self.node = node
        self.col = col

class Solution:
    """
    @param root: the root of tree
    @return: the vertical order traversal
    """
    def verticalOrder(self, root):
        res = []
        if not root:
            return res
        queue = [TreeColumnNode(root, 0)]
        cur_num, nxt_num = 1, 0
        min_value, max_value = 0, 0
        res_dict = {}
        while queue:
            cur_node = queue.pop()
            cur_num -= 1
            if cur_node.col in res_dict:
                res_dict[cur_node.col].append(cur_node.node.val)
            else:
                res_dict[cur_node.col] = [cur_node.node.val]
            if cur_node.node.left:
                queue.insert(0, TreeColumnNode(cur_node.node.left, cur_node.col - 1))
                nxt_num += 1
                min_value = min(min_value, cur_node.col - 1)
            if cur_node.node.right:
                queue.insert(0, TreeColumnNode(cur_node.node.right, cur_node.col + 1))
                nxt_num += 1
                max_value = max(max_value, cur_node.col + 1)
            if cur_num == 0:
                cur_num = nxt_num
                nxt_num = 0
        for i in range(min_value, max_value + 1):
            res.append(res_dict[i])
        return res