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,null,null,15,7],
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
思路： 逐个level操作。用queue来维护每一行，对于每一行pop出第一个node的时候，把他的left和right kid push到queue里面。注意应该先算出这一行到底有多少个node （也就是当前queue的长度），否则不知道什么时候该停下来。
1 # Definition for a binary tree node. 2 # class TreeNode(object): 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution(object): 9 def levelOrder(self, root): 10 """ 11 :type root: TreeNode 12 :rtype: List[List[int]] 13 """ 14 if not root: 15 return [] 16 17 ans = [] 18 queue = [root] 19 while queue: 20 level = [] 21 size = len(queue) 22 for i in range(size): 23 node = queue.pop(0) 24 level.append(node.val) 25 if node.left: 26 queue.append(node.left) 27 if node.right: 28 queue.append(node.right) 29 ans.append(level) 30 return ans
Leetcode 102. Binary Tree Level Order Traversal
原文：http://www.cnblogs.com/lettuan/p/6412075.html