102. Binary Tree Level Order Traversal

Not much to say. BFS can be applied here. Here is the code.

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:  public:  
12:    vector<vector<int>> levelOrder(TreeNode* root) {  
13:      vector<vector<int>> res;  
14:      if (root == NULL) return res;  
15:      queue<TreeNode*> q;  
16:      q.push(root);  
17:      while (!q.empty()) {  
18:        int sz = q.size();  
19:        vector<int> level;  
20:        for (int i = 0; i < sz; i++) {  
21:          TreeNode *node = q.front();  
22:          q.pop();  
23:          level.push_back(node->val);  
24:          if (node->left) q.push(node->left);  
25:          if (node->right) q.push(node->right);  
26:        }  
27:        res.push_back(level);  
28:      }  
29:      return res;  
30:    }  
31:  };