LeetCode OJ：Binary Tree Level Order Traversal II（二叉树的层序遍历）

 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 private:
12     struct Node{
13         TreeNode * treeNode;
14         int level;
15         Node(){}
16         Node(TreeNode * nd, int lv)
17         :treeNode(nd), level(lv){}
18     };
19 public:
20     vector<vector<int>> levelOrderBottom(TreeNode* root) {
21         vector<vector<int>> ret;
22         if(!root)
23             return ret;
24         queue<Node> nodeQueue;
25         nodeQueue.push(Node(root, 0));
26         int dep = -1;
27         while(!nodeQueue.empty()){
28             Node node = nodeQueue.front();
29             if(node.treeNode->left)
30                 nodeQueue.push(Node(node.treeNode->left, node.level + 1));
31             if(node.treeNode->right)
32                 nodeQueue.push(Node(node.treeNode->right, node.level + 1));
33             if(dep == node.level)
34                 ret[dep].push_back(node.treeNode->val);
35             else{
36                 vector<int> tmp;
37                 dep++;
38                 ret.push_back(tmp);
39                 ret[dep].push_back(node.treeNode->val);
40             }
41             nodeQueue.pop();    //不要忘了
42         }
43         reverse(ret.begin(), ret.end());
44         return ret;
45     }
46 };