【LeetCode with Python】 Binary Tree Level Order Traversal II

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原题页面：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
题目类型：
难度评价：★
本文地址：http://blog.csdn.net/nerv3x3/article/details/37329485

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

不知道出题意图是不是希望用递归，从叶子节点到根结点，将每一个节点加入到结果二维数组的相应层次中？
不过我这里就追求简单，直接将Binary Tree Level Order Traversal的结果数组reverse一下。。。

class Solution:    # @param root, a tree node    # @return a list of lists of integers    def levelOrder(self, root):        if None == root:            return [ ]        results = [ [root.val] ]        reflist1 = [root]        while True:            reflist2 = [ ]            result = [ ]            for i in range(0, len(reflist1)):                cur = reflist1[i]                if None != cur.left:                    reflist2.append(cur.left)                    result.append(cur.left.val)                if None != cur.right:                    reflist2.append(cur.right)                    result.append(cur.right.val)            if 0 == len(reflist2):                break            results.append(result)            reflist1 = reflist2        return results    def levelOrderBottom(self, root):        results = self.levelOrder(root)        results.reverse()        return results