102 Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

/**  * Definition for binary tree  * struct TreeNode {  *     int val;  *     TreeNode *left;  *     TreeNode *right;  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}  * };  */ class Solution { public:     vector<vector<int> > levelOrder(TreeNode *root) {         vector<vector<int> > result;         if (root == NULL) return result;                   queue<TreeNode*> q;         q.push(root);                   while (!q.empty()) {             int count = q.size();             result.push_back(vector<int>());             for (int i = 0; i < count; i++) {                 TreeNode *tmp = q.front();                 q.pop();                 if (tmp->left != NULL) q.push(tmp->left);                 if (tmp->right != NULL) q.push(tmp->right);                 result.back().push_back(tmp->val);             }         }                   return result;            } };