标签:
Basic question. Use a queue to store Node, use two numbers to record current level node numbers and next level node numbers.
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> > levelOrder(TreeNode *root) { 13 vector<vector<int> > result; 14 vector<int> level; 15 if (!root) return result; 16 int current = 1, future = 0; 17 queue<TreeNode *> q; 18 q.push(root); 19 while(!q.empty()) { 20 TreeNode *tmp = q.front(); 21 q.pop(); 22 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 result.push_back(level); 36 level.clear(); 37 } 38 } 39 return result; 40 } 41 };
LeetCode – Refresh – Binary Tree Level Order Traversal
标签:
原文地址:http://www.cnblogs.com/shuashuashua/p/4346173.html