LeetCode: 107_Binary Tree Level Order Traversal II | 二叉树自底向上的层次遍历 | Easy

 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 vector<vector<int> > levelOrderTraversalII(TreeNode *root) //非递归的中序遍历(用栈实现)
10 {
11 
12     queue<TreeNode *> tree_queue;
13     vector<vector<int> > tree_vector, tree_rvector;
14     vector<int> svector;
15     int nCount = 0;
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             nCount ++;
31             svector.clear();
32             if (tree_queue.empty())
33                 break;
34             tree_queue.push(NULL);
35         }
36         else {
37             svector.push_back(pTemp->val);
38             if (pTemp->left) 
39                 tree_queue.push(pTemp->left);
40             if (pTemp->right)
41                 tree_queue.push(pTemp->right);
42         }
43     }
44     while (nCount > 0){
45         svector = tree_vector.at(nCount - 1);
46         tree_rvector.push_back(svector);
47         nCount --;
48     }
49     return tree_rvector;
50 }