题目描述：
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}".
代码：
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int> > res;
vector<int> lev;
if(root == NULL) return res;
queue<TreeNode *> que;
que.push(root);
que.push(NULL); //end of one level
while(true)
{
TreeNode *cur = que.front();
que.pop();
if(!cur)
{
res.push_back(lev);
lev.clear();
if(que.empty()) break;
que.push(NULL);
}
else
{
lev.push_back(cur->val);
if(cur->left) que.push(cur->left);
if(cur->right) que.push(cur->right);
}
}
return res;
}
};