116. 填充每个节点的下一个右侧节点指针

1. 题目链接

2. 解题思路：其实就是在层序遍历的过程中，把next指针填上

3. 代码

   class Solution:
       def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
           if root == None:
               return None
           # 层序遍历的过程中，填好next
           queue = deque()
           queue.append(root)
           cur_count = 1
           next_count = 0
           while queue:
               cur = queue.popleft()
               if cur.left:
                   next_count += 1
                   queue.append(cur.left)
               if cur.right:
                   next_count += 1
                   queue.append(cur.right)
               cur_count -= 1
               if cur_count == 0:
                   cur_count = next_count
                   next_count = 0
               else:
                   cur.next = queue[0]
           return root