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;  }}
};