leetcode日记（101）填充每个节点的下一个右侧节点指针Ⅱ

意料之中有这题，将之前的思路换一下即可，层序遍历的思路将record（记录下一个循环的次数）手动加减。

/*
// 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:
    Node* connect(Node* root) {
        if(root==NULL) return root;
        queue<Node*> q;
        q.push(root);
        int frequency=1;
        while(!q.empty()){
            int record=0;
            Node* n=NULL;
            for(int i=0;i<frequency;i++){
                Node* first=q.front();
                if(first->left){
                    q.push(first->left);
                    record++;
                }
                if(first->right){
                    q.push(first->right);
                    record++;
                }
                if(n) n->next=first;
                n=first;
                q.pop();
                record--;
            }
            frequency+=record;
        }
        return root;
    }
};

递归方法也差不多，只要"稍改"一点即可。

/*
// 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:
    Node* connect(Node* root) {
        Node* result=root;
        while(root){
            Node* record=root->left?root->left:root->right;
            Node* last=NULL;
            Node* next=NULL;
            while(root){
                if(root->right&&root->left){root->left->next=root->right;}
                else if(root->right==NULL&&root->left==NULL){
                    root=root->next;
                    if(record==NULL&&root) record=root->left?root->left:root->right;
                    continue;
                }
                next=root->left?root->left:root->right;
                if(last&&next) last->next=next;
                last=root->right?root->right:root->left;
                root=root->next;
            }
            root=record;
        }
        return result;
    }
};

递归的很容易出错，还是用前面那种方法更好。