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