leetcode 102

102. Binary Tree Level Order Traversal

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]
]

层次遍历二叉树，并要求将每一层的数放在二维数组的同一行中。思路：采用两个队列，队列1存放上一层结点，队列1中的结点一次出队并在队列2存放下一层结点，另外用一个数组记录队列1中结点的值，这些值就属于一层。　　　将数组中的值放入二维数组相应的行，并清空数组。队列2中的结点出队并一次放入队列1中。

代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         vector<vector<int>> lorder;
14         if(root == NULL)
15         {
16             return lorder;
17         }
18         queue<TreeNode*> q;
19         q.push(root);
20         vector<int> level_tmp2;
21         while(!q.empty())
22         {
23             level_tmp2.clear();
24             queue<TreeNode*> level_tmp1;
25             TreeNode* node = q.front();
26             int size = q.size();
27
28             for(int i = 0; i < size; i++)
29             {
30                 TreeNode* node = q.front();
31                 q.pop();
32                 if(node->left)
33                 {
34                     level_tmp1.push(node->left);
35                 }
36                 if(node->right)
37                 {
38                     level_tmp1.push(node->right);
39                 }
40                 level_tmp2.push_back(node->val);
41             }
42             while(!level_tmp1.empty())
43             {
44                 q.push(level_tmp1.front());
45                 level_tmp1.pop();
46             }
47             lorder.push_back(level_tmp2);
48         }
49        return lorder;
50     }
51 };