LeetCode: 102_Binary Tree Level Order Traversal | 二叉树自顶向下的层次遍历 | Easy

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},

 1 struct TreeNode {
 2     int        val;
 3     TreeNode*    left;
 4     TreeNode*    right;
 5     TreeNode():val(0),left(NULL),right(NULL) {}
 6     TreeNode(int x): val(x), left(NULL),right(NULL) {}
 7 };
 8 
 9 
10 vector<vector<int> > levelOrderTraversal(TreeNode *root) //非递归的中序遍历(用栈实现)
11 {
12 
13     queue<TreeNode *> tree_queue;
14     vector<vector<int> > tree_vector;
15     vector<int> svector;
16 
17     if (NULL == root) {
18         return tree_vector;
19     }
20     TreeNode *pTemp = root;
21     tree_queue.push(root);
22     tree_queue.push(NULL); //the end of one level.
23 
24     while (true) {
25         TreeNode *pTemp = tree_queue.front();
26         tree_queue.pop();
27 
28         if (!pTemp) { //get the null, put vector<> to vector<vector<>>
29             tree_vector.push_back(svector);
30             svector.clear();
31             if (tree_queue.empty())
32                 break;
33             tree_queue.push(NULL);
34         }
35         else {
36             svector.push_back(pTemp->val);
37             if (pTemp->left) 
38                 tree_queue.push(pTemp->left);
39             if (pTemp->right)
40                 tree_queue.push(pTemp->right);
41         }
42     }
43     return tree_vector;
44 }