给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

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

这个题要充分利用完美二叉树的特点，递归的方法就是逐个子树进行连接，当前节点存在左右子树，那么左节点一定可以链接到右节点，然后判断当前节点的next是否存在，如果存在，那么右节点可以链接到next节点的左节点。然后依次递归。迭代的方法非常的厉害，利用一个首节点和当前节点来遍历整层节点，首节点就是不断的搜索左节点，左节点存在则把首节点赋给当前节点，然后当前节点的左节点链接到右节点，判断当前节点的next是否存在，存在则把右节点链接到next的左节点，然后当前节点变为next接续遍历。

C++源代码：

/*
// 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) return NULL;
        if(root->left) root->left->next = root->right;
        if(root->right) root->right->next = root->next?root->next->left:NULL;
        connect(root->left);
        connect(root->right);
        return root;
    }
};

python3源代码：

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if root==None:
            return None
        start = root
        cur = None
        while start.left:
            cur = start
            while cur:
                cur.left.next = cur.right
                if cur.next:
                    cur.right.next = cur.next.left
                cur = cur.next
            start = start.left
        return root