Binary Tree Level Order Traversal---102--二叉樹層序遍歷

題目描述：給出一顆二叉樹，返回他的層序遍歷，隊列的解法

input 1，null， 2，3   output {[1]，[2]，[3]}

思路：層序遍歷是廣度優先的BFS的應用
首先建立一個隊列，根節點入隊
隊列不爲空的情況下，根節點出隊，左右子節點入隊，此時隊列裏是下一層的節點
for循環遍歷這一層的隊列，同時再把他們的子節點入隊，上一層的節點出隊，依次類推就是層序遍歷

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct TreeNode
{
    int data;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : data(x), left(NULL), right(NULL) {}
};
TreeNode *creatTree(vector<int> &v)
{
    TreeNode *root;
    if (v.size() == 0)
        return NULL;
    for (int i = 0; i < v.size(); ++i)
    {
        if (v[0] == -1)
        {
            root = NULL;
            v.erase(v.begin());
        }
        else
        {
            root = new TreeNode(v[0]);
            v.erase(v.begin());
            root->left = NULL;
            root->left = creatTree(v);
            root->right = NULL;
            root->right = creatTree(v);
        }
    }
    return root;
}

vector<vector<int> > levelOrder(TreeNode *root)
{
   if(!root) return {};
   vector<vector<int> > res;//二維變量存儲
   queue<TreeNode*> q;
   q.push(root);
   while(!q.empty())//隊列不爲空的時候
   {
        vector<int> oneLevel;    //一維存儲每一層的節點
        for(int i = q.size(); i > 0; --i)
        {
            TreeNode* temp = q.front();
            q.pop();
            oneLevel.push_back(temp->data);//隊首節點出隊
            if(temp->left) q.push(temp->left);
            if(temp->right) q.push(temp->right);//左右子節點分別入隊
        }
        res.push_back(oneLevel);
   }
   return res;
}
int main()
{
    vector<int> v;
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i)
    {
        int data;
        cin >> data;
        v.push_back(data);
    }
    TreeNode *T = creatTree(v);
    vector<vector<int> > res;
    res = levelOrder(T);
    return 0;
}