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Given a binary tree

struct Node {  int val;  Node *left;  Node *right;  Node *next;}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

用层序遍历的方式，把每一层节点存在一个list中。

然后进行处理比较容易。

class Solution:    def connect(self, root: 'Node') -> 'Node':        if root is None:return root        level = [root]        root.next=None        while level:            temp=[]            for i in range(len(level)-1):                level[i].next=level[i+1]            for node in level:                if node.left:                    temp.append(node.left)                if node.right:                    temp.append(node.right)            level=temp        return root

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