【剑指offer】59 之字形打印树

题目

分析

bfs遍历树，同时当count%2等于0时正向打印，count%2等于1时逆向打印

python代码

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def Print(self, pRoot):
        # write code here
        if not pRoot: return []
        res = []
        arr = [pRoot]
        count = 0
        while (len(arr)!=0):
            temp_arr = []
            temp_res = []
            for n in arr:
                if n:
                    temp_res.append(n.val)
                    if n.left:
                        temp_arr.append(n.left)
                    if n.right:
                        temp_arr.append(n.right)
            if count%2==0:
                res.append(temp_res)
            else:
                res.append(temp_res[::-1])
            arr = temp_arr
            count+=1
        return res