题目描述：

给定一个二叉树

struct Node {
int val;
Node left;
Node right;
Node *next;
}
填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

提示：

• 你只能使用常量级额外空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

解法：

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

    Node() {}

    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;
        }else{
            Node* head = root;
            Node* pre = NULL;
            Node* cur = head;
            while(head){
                cur = head;
                head = NULL;
                pre = NULL;
                while(cur){
                    if(cur->left){
                        if(head == NULL){
                            head = cur->left;
                        }
                        cur->left->next = cur->right;
                        if(pre == NULL){
                            pre = cur->left;
                        }else{
                            pre->next = cur->left;
                            pre = cur->left;
                        }
                    }
                    
                    if(cur->right){
                        if(head == NULL){
                            head = cur->right;
                        }
                        if(pre == NULL){
                            pre = cur->right;
                        }else{
                            pre->next = cur->right;
                            pre = cur->right;
                        }
                    }
                    cur = cur->next;
                }
            }
            return root;  
        }
    }
};