LeetCode 117. 填充每个节点的下一个右侧节点指针 II

原题链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

题目分析

至于为什么先右子树，如果左子树优先的话，root->right需要通过root的next一直找，直到找到同层最右边的节点，此时同层右子树的节点还没连接

C++代码

#include
using namespace std;


// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};

class Solution {
public:

    bool update(Node *origin,Node *temp){
        bool flag = true;
        if(temp->left){
            origin->next = temp->left;
        }else if(temp->right){
            origin->next = temp->right;
        }else{
            origin->next = nullptr;
            flag = false;
        }
        return flag;
    }
    Node* connect(Node* root) {
        if (root == nullptr) return nullptr;    
       // cout<val<<" ";
        if(root->left){
            if(root->right) root->left->next = root->right;
            else{
                Node* temp = root->next;
                while( temp && !update(root->left,temp)){
                    temp = temp->next;
                }
            }
        }
        if (root->right) {
            Node* temp = root->next;
            while( temp && !update(root->right,temp)){
                cout<val<<" ";
                if(temp->next) cout<next->val<next;
            }
            if(root->next == nullptr){
                root->right->next = nullptr;
            }
        }
        root->right = connect(root->right);
        root->left = connect(root->left);
        return root;
    }
};