zoukankan      html  css  js  c++  java
  • 116. Populating Next Right Pointers in Each Node 连接右节点

    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.

     

    Follow up:

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

     

    Example 1:

    Input: root = [1,2,3,4,5,6,7]
    Output: [1,#,2,3,#,4,5,6,7,#]
    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. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level

    思路:traverse的递归。左边的不为空,才能连接到右边
    return;表示树中的递归结束了,不需要返回任何东西

    class Solution {
        public void connect(Node root) {
            //cc
            if (root == null)
                return ;
            
            //连接
            if (root.left != null) {
               root.left.next = root.right;
                    
                if (root.right != null) {
                    root.right.next = root.next;
                }
            }
                
            //左右connect
            connect(root.left);
            connect(root.right);
        }
    }
    View Code
    .
     
  • 相关阅读:
    进程和线程的概述
    注意两个词汇的区别:并行和并发
    WebRTC MCU( Multipoint Conferencing Unit)服务器调研
    (译)WebRTC实战: STUN, TURN, Signaling
    关于图数据库查询语言:Cypher
    Neo4j安装后的密码修改
    XYC2016上半年工作笔记整理
    WebRTC技术调研
    在Django中使用Neo4j
    传统企业做互联网的困局
  • 原文地址:https://www.cnblogs.com/immiao0319/p/13882171.html
Copyright © 2011-2022 走看看