层序遍历I：单层vector
思想：
用队列控制前后，用size来控制每一层
解法：
class Solution {
public:
vector<int> levelOrder(TreeNode* root) {
if(root == NULL)
return {
};
vector<int>res;
queue<TreeNode*>my_queue;
my_queue.push(root);
while(!my_queue.empty()){
int size = my_queue.size();
//获取每层的个数，只遍历当前层
while(size--){
TreeNode* cur = my_queue.front();
my_queue.pop();
res.push_back(cur->val);
if(cur->left)
my_queue.push(cur->left);
if(cur->right)
my_queue.push(cur->right);
}
}
return res;
}
};
层序遍历II：双层vector
代码：
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if(root == NULL)
return {
};
vector<vector<int>>res;
queue<TreeNode*>my_queue;
my_queue.push(root);
while(!my_queue.empty()){
int size = my_queue.size();
//每层开始时创建一个tmp，结束时将它插入res即可
vector<int>tmp;
//获取每层的个数，只遍历当前层
while(size--){
TreeNode* cur = my_queue.front();
my_queue.pop();
tmp.push_back(cur->val);
if(cur->left)
my_queue.push(cur->left);
if(cur->right)
my_queue.push(cur->right);
}
res.push_back(tmp);
}
return res;
}
};
层序遍历III：之字型双层vector
解法1：迭代器
思想：
●用一个bool型变量reverse来记录是否需要反转，每层reverse取反一次，注意reverse要定义在循环外面
●用vector(tmp.rbegin(), tmp.rend())来获得tmp的倒序，也可以用reverse(tmp.begin(), tmp.end())
代码：
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if(root == NULL)
return {
};
vector<vector<int>>res;
queue<TreeNode*>my_queue;
my_queue.push(root);
//reverse标志位，每一层改变一次，注意要在while的外面定义
bool reverse = false;
while(!my_queue.empty()){
int size = my_queue.size();
//每层开始时创建一个tmp，结束时将它插入res即可
vector<int>tmp;
//获取每层的个数，只遍历当前层
while(size--){
TreeNode* cur = my_queue.front();
my_queue.pop();
tmp.push_back(cur->val);
if(cur->left)
my_queue.push(cur->left);
if(cur->right)
my_queue.push(cur->right);
}
if(reverse)
//tmp.rbegin()和tmp.rend()来倒着插入
res.push_back(vector<int>(tmp.rbegin(), tmp.rend()));
else
res.push_back(tmp);
reverse = !reverse;
}
return res;
}
};