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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

其實就是樹的廣度優先搜尋，使用佇列進行輔助，遍歷每一層
/**
 * Definition for a binary tree node.
 * 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>> res; 
        if (root == NULL) return res;
        vector<int> sub;    // 儲存每一層每個結點的值

        queue<TreeNode *> q; // 輔助佇列
        q.push(root);

        while
 
 (!q.empty()) { // q為空時結束迴圈
            queue<TreeNode *> tmpq; // 臨時佇列，儲存下一層的結點
            while (!q.empty()) {
                sub.push_back(q.front()->val);  // 將該層節點值加入sub中
                if (q.front()->left != NULL) tmpq.push(q.front()->left); // 依次將左右子結點放入佇列中
                if (q.front()->right != NULL) tmpq.push(q.front()->right);
                q.pop(); // 彈出一個結點，為空時表明當前層的所有結點都被遍歷過
 

            }
            res.push_back(sub); // 加入結果集
            q = tmpq;         // 更新佇列
            sub.clear();      // 清空，以備下次使用
        }
        return res;
    }
};