LeetCode102. 二叉树的层序遍历

题目要求是在输出中每层分开存放——对于该要求，有一个巧妙的处理方式：在循环处理每层节点之前，固定该层的节点数，这样在处理完该层节点后就自动停止

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
     vector<vector<int>> levelOrder(TreeNode* root) 
    {   
        TreeNode* cur;
        deque<TreeNode*> queue;
        vector<vector<int>> res;
        if (root == NULL) return res;
        /* 根节点入队 */
        queue.push_back(root);
        /* 遍历 */
        while (!queue.empty())
        {
            int size = queue.size();
            vector<int> curLevel;
            for (int i = 0; i < size; i++)
            {
                cur = queue.front();
                queue.pop_front();
                curLevel.push_back(cur->val);
                if (cur->left != NULL)
                    queue.push_back(cur->left);
                if (cur->right != NULL)
                    queue.push_back(cur->right);
            }
            res.push_back(curLevel);    
        }
        return res;
    }
};