LeetCode – Refresh – Binary Tree Level Order Traversal ii-白红宇

LeetCode – Refresh – Binary Tree Level Order Traversal ii

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发布时间：2019-06-22

本文共 1240 字，大约阅读时间需要 4 分钟。

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
     
      
        > levelOrderBottom(TreeNode *root) {13         vector
       
        
          > result;14         if (!root) return result;15         stack
         
          
            > s;16 queue
           
             q;17 vector
            
              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 };