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填充每个节点的下一个右侧节点指针 II

Populating Next Right Pointers in Each Node II

专题
Algorithmic Programming / 算法编程
难度
L3
来源
Citadel

题目详情

问题:填充每个节点的下一个右侧节点指针 II

考察:树

来源:Citadel

链接:https://www.jointaro.com/interviews/questions/populating-next-right-pointers-in-each-node-ii/

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.

 

Example 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

 

Constraints:

  • The number of nodes in the tree is in the range [0, 6000].
  • -100 <= Node.val <= 100

 

Follow-up:

  • You may only use constant extra space.
  • The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.
解析

思路:普通二叉树不能假设子节点完整。逐层遍历当前层 next 链,用 dummy 指针串起下一层所有孩子,处理完后进入 dummy.next。

复杂度:时间 O(n),额外空间 O(1)。