problem：
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.
thinking：
（1）普通的二叉树层序遍历算法每层是没有分开的，这道题要求分开。
（2）采用广度优先搜索（BFS）的思想
（3）采用两个queue，递归实现每层分开的层序遍历，很经典
code：
class Solution {
private:
vector<vector<int> > ret;
public:
vector<vector<int> > levelOrder(TreeNode *root) {
ret.clear();
if(root==NULL)
return ret;
queue<TreeNode *> tmp_queue;
tmp_queue.push(root);
level_order(tmp_queue);
return ret;
}
void level_order(queue<TreeNode *> queue1)
{
if(queue1.empty())
return;
vector<int> array;
queue<TreeNode *> queue2;
while(!queue1.empty())
{
TreeNode *tmp=queue1.front();
array.push_back(tmp->val);
queue1.pop();
if(tmp->left!=NULL)
queue2.push(tmp->left);
if(tmp->right!=NULL)
queue2.push(tmp->right);
}
ret.push_back(array);
level_order(queue2);
}
};leetcode || 102、Binary Tree Level Order Traversal
原文地址：http://blog.csdn.net/hustyangju/article/details/45147231
