leetcode 117. Populating Next Right Pointers in Each Node II 給非完全二叉樹每個節點增加一個水平指針

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

1.遞歸-對每層新建一個外部節點，用於將每層串起來。由於所有節點的next初始化爲null，因此對於最右側的節點不作處理。

/*
// 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==nullptr) return root;
        Node out(-1);
        for(Node* cur=root, *p=&out; cur!=nullptr;cur=cur->next){
            if(cur->left!=nullptr){
                p->next = cur->left;
                p = p->next;
            }
            if(cur->right!=nullptr){
                p->next = cur->right;
                p = p->next;
            }
        }
        connect(out.next);
        return root;
    }
};

2.非遞歸-雙隊列法

/*
// 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==nullptr) return root;
        queue<Node*> a, b;
        a.push(root);
        while(!a.empty() || !b.empty()){
            while(!a.empty()){
                Node* tmp = a.front();
                a.pop();
                if(a.empty())
                    tmp->next = nullptr;
                else
                    tmp->next = a.front();
                if(tmp->left!=nullptr)
                    b.push(tmp->left);
                if(tmp->right!=nullptr)
                    b.push(tmp->right); 
            }
            
            while(!b.empty()){
                Node* tmp = b.front();
                b.pop();
                if(b.empty())
                    tmp->next = nullptr;
                else
                    tmp->next = b.front();
                if(tmp->left!=nullptr)
                    a.push(tmp->left);
                if(tmp->right!=nullptr)
                    a.push(tmp->right);  
            }
        }
        return root;
    }
};