LeetCode 문제: Populating Next Right Pointers in Each Node

1734 단어 LeetCode
Given a binary tree
struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *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.
Note:
You may only use constant extra space. You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children). For example, Given the following perfect binary tree, 1 /\ 2 3 /\/\ 4 5 6 7 After calling your function, the tree should look like: 1 -> NULL /\ 2 -> 3 -> NULL /\/\ 4->5->6->7 -> NULL
제목: 두 갈래 나무의 각 층에 양보하는 결점은 왼쪽에서 오른쪽으로 그 오른쪽 결점을 가리키고, 가장 오른쪽의 결점은 NULL을 가리킨다
해결 사고방식: 층층이 두루 훑어보다
코드:
public class Solution {
    public void connect(TreeLinkNode root) {
       while (root != null) {
           TreeLinkNode curr = root;

           while (curr != null && curr.left != null) {
               curr.left.next = curr.right;
               curr.right.next = curr.next == null ? null : curr.next.left;
               curr = curr.next;
           }

           root = root.left;
       }
    }
}

좋은 웹페이지 즐겨찾기