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.
- 类似 https://www.cnblogs.com/pegasus923/p/11190990.html 。BFS一层层扩展,把子节点加到下一层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