题目简述：

填充它的每个next指针，让这个指针指向其下一个右侧节点。如果找不到则为null

class Solution {
public:
    Node* connect(Node* root) {
        if(root == nullptr) return root;
        deque<Node*> rel;
        rel.push_back(root); // 队列每次都新增一个元素
        int level_size = 1; // 某层的容量
        while(!rel.empty()) {
            if(rel.size() == level_size) {
                for(int i = 0; i < level_size-1; i++) {
                    rel.at(i)->next = rel.at(i+1);
                }
                level_size *= 2; 
            }
            Node* front = rel.at(0);
            cout<<front->val<<endl;
            rel.pop_front();
            if(front->left != nullptr) rel.push_back(front->left);
            if(front->right != nullptr) rel.push_back(front->right);
        }
        return root;
    }
};

复习二叉树的层序遍历

#include<cstdio>
#include<queue>
using namespace std;
struct BinaryTree{
    int vec;
    BinaryTree* left;
    BinaryTree* right;
    BinaryTree(int data) : vec(data), left(nullptr), right(nullptr) {}
};

// 队列实现层序遍历
void printTree(BinaryTree* arr[]) {
    queue<BinaryTree*> rel; // 定义一个队列，数据类型是二叉树指针
    res.push(arr[0]);
    while(!rel.empty()) {
        BinaryTree* front = rel.front;
        printf("%d\n", front->vec);
        rel.pop(); // 删除最前面的节点
        if(front->left != nullptr) rel.push(front->left);
        if(front->right != nullptr) rel.push(front->right);
    }
}