LeetCode – Refresh – Binary Tree Level Order Traversal ii

This question is almost same as the previous one.

I just use a stack (another vector) of vector<int> to store the level result. Then reverse it.

 1 /**
 2  * Definition for binary tree
 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> > levelOrderBottom(TreeNode *root) {
13         vector<vector<int> > result;
14         if (!root) return result;
15         stack<vector<int> > s;
16         queue<TreeNode *> q;
17         vector<int> level;
18         q.push(root);
19         int current = 1, future = 0;
20         while (!q.empty()) {
21             TreeNode *tmp = q.front();
22             q.pop(), current--;
23             if (tmp->left) {
24                 q.push(tmp->left);
25                 future++;
26             }
27             if (tmp->right) {
28                 q.push(tmp->right);
29                 future++;
30             }
31             level.push_back(tmp->val);
32             if (!current) {
33                 current = future;
34                 future = 0;
35                 s.push(level);
36                 level.clear();
37             }
38         }
39         while (!s.empty()) {
40             result.push_back(s.top());
41             s.pop();
42         }
43         return result;
44     }
45 };

 

转载于:https://www.cnblogs.com/shuashuashua/p/4346177.html