Leetcode 116. Populating Next Right Pointers in Each Node

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

Medium

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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.

 

Example:

Input: {"$id":"1","left":{"$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}

Output: {"$id":"1","left":{"$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}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

 

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

---

• 相似 http://www.javashuo.com/article/p-eazmlxar-kr.html 。BFS一層層擴展，把子節點加到下一層list中。

 1 """
 2 # Definition for a Node.
 3 class Node:
 4     def __init__(self, val, left, right, next):
 5         self.val = val
 6         self.left = left
 7         self.right = right
 8         self.next = next
 9 """
10 class Solution:
11     def connect(self, root: 'Node') -> 'Node':
12         if root is None:
13             return None
14         
15         head = root
16         list_current = [root]
17         
18         while list_current:
19             next_level = []
20             
21             for i in range( len( list_current ) ):
22                 if i + 1 < len(list_current):
23                     list_current[i].next = list_current[i + 1]
24                 
25                 if list_current[i].left:
26                     next_level.append(list_current[i].left)
27                 if list_current[i].right:
28                     next_level.append(list_current[i].right)
29             
30             list_current = next_level
31         
32         return head            

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