[LeetCode] Binary Tree Level Order Traversal

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

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

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

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

这题目其实是要求按层遍历二叉树，并且分层输出。在遍历二叉树时最简单就是递归了，但简单递归无法保存层的信息，在借助队列遍历时，也只能简单的按层输出，无法确

元素在那一层，这里借鉴了别人的思想：设置两个标记量分别标记parent数和child数，再说一次，要得到按层遍历的序列不难，难的是怎么确定什么时候是分层，开始的

时候queue只有root一个元素，此时parentIndex = 1, childIndex = 0，由于它的left和right都为飞空，所以root.left 和 root.right 分别入队列,

childIndex 递增两次，即childIndex += 2。。。算了，我也说不清，还是上代码吧：

class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Noneclass Solution:# @param root, a tree node# @param sum, an integer# @return a booleandef levelOrder(self, root):if None == root:return []queue = [root]ret = []tmp = []parentIndex = 1childIndex = 0while len(queue) > 0:leaf = queue[0]tmp.append(leaf.val)del queue[0]if None != leaf.left:queue.append(leaf.left)childIndex += 1if None != leaf.right:queue.append(leaf.right)childIndex += 1parentIndex -= 1if 0 == parentIndex:ret.append(tmp)tmp = []parentIndex = childIndexchildIndex = 0return ret

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