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]
]

题意：对于给定的二叉树，返回按层序遍历的结点值

思路：利用队列，BFS

/**
 * 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>> ans;
        queue<TreeNode *> que;
        if(!root) return ans;
        que.push(root);
        while(!que.empty()){
            vector<int> temp;  //一层地元素
            for(int i=que.size();i>0;i--) 
            {
                TreeNode *t=que.front();
                que.pop();
                temp.push_back(t->val);
                if(t->left) que.push(t->left);
                if(t->right) que.push(t->right);
            }
            ans.push_back(temp);      //插入到ans数组中
        }
        return ans;
    }
};