涉及C++ STL容器的强制类型转换

用C++做leetcode的时候偶有出现函数返回类型需要从deque转到vector的情况，之前不熟悉相关语法，故做个记录备忘。
注意看倒数第56行代码的强制类型转换细节，用的是花括号！

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

#include <iostream>
#include <deque>
#include <algorithm>
using namespace std;

class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode* root) {

        vector<vector<int> > ans;

        if (root == nullptr ) return ans;

        deque<TreeNode*> deq;

        int round = 0;

        deq.push_back(root);

        while (!deq.empty() ) {

            deque<int> tempDeque;

            int size = deq.size();

            for (int i = 0; i < size; ++i) {

                TreeNode* tempNode = deq.front();

                deq.pop_front();

                if (round % 2 == 0)
                    tempDeque.push_back(tempNode->val);
                else
                    tempDeque.push_front(tempNode->val); //关键代码！！

                if (tempNode->left != nullptr) deq.push_back(tempNode->left);

                if (tempNode->right != nullptr) deq.push_back(tempNode->right);

            }

            ++round;

            //强制类型转换
            ans.push_back(vector<int>{tempDeque.begin(), tempDeque.end() } );

        }


        return ans;
        
    }
};