zoukankan      html  css  js  c++  java
  • [leetcode]117. Populating Next Right Pointers in Each NodeII用next填充同层相邻节点

    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.
    • Recursive approach is fine, implicit stack space does not count as extra space for this problem.

    Example:

    Given the following binary tree,

         1
       /  
      2    3
     /     
    4   5    7

    After calling your function, the tree should look like:

         1 -> NULL
       /  
      2 -> 3 -> NULL
     /     
    4-> 5 -> 7 -> NULL

    题意:

    给定一个任意二叉树,其中每个节点都有next指针

    设法将next指针指向同一层右侧相邻节点

    思路:

    递归,解法很巧妙。

    以 1 为curr,用pre来连接curr.left:2 和 curr.right:3。因为此时根节点 1的next指向null, 退出for循环
                1  ( curr )--->null 
                /  
    dummy(-1):pre--> 2---> 3
               /     
              4   5    7




    递归调用connect(dummy.next),因为dummy.next指向2此时 2 为curr。用pre来连接curr.left:4 和 curr.right:5。
                       1  
                   /        
                       2(curr)---> 3
                 /             
    dummy(-1):pre--> 4--->5 7

    继续走for循环, curr = curr.next, 此时curr为3。 继续用pre连接 curr.left(3的左子树为Null) 和 curr.right:7。

          1
                 /    
                       2 ---> 3(curr)
             /
    dummy(-1):pre--> 4--->5 --->7

    因为此时curr的next指向null, 退出for循环。
    继续递归调用connect(dummy.next)... 直至 root == null, return

    代码:

     1 public class Solution {
     2     public void connect(TreeLinkNode root) {
     3         if (root == null) return;
     4 
     5         TreeLinkNode dummy = new TreeLinkNode(-1);
     6         for (TreeLinkNode curr = root, prev = dummy;
     7                 curr != null; curr = curr.next) {
     8             if (curr.left != null){
     9                 prev.next = curr.left;
    10                 prev = prev.next;
    11             }
    12             if (curr.right != null){
    13                 prev.next = curr.right;
    14                 prev = prev.next;
    15             }
    16         }
    17         connect(dummy.next);
    18     }
    19 }
  • 相关阅读:
    ASP.NET里面,如果设置了form的 onsubmit="return false;"之后,就不能提交按钮了?
    存储过程(待填)
    关于电脑屏幕分辨率太高字太小怎么解决?
    最新版Android开发工具
    Xamarin For Visual Studio 3.0.54.0 完整离线破解版(C# 开发Android、IOS工具 吾乐吧软件站分享)
    ubuntu 16.04 source (HUST and 163)
    transmission简单使用
    Test Tex
    What is a Statistic?
    IDE、SATA、SCSI、SAS、FC、SSD硬盘类型介绍[zz]
  • 原文地址:https://www.cnblogs.com/liuliu5151/p/9124367.html
Copyright © 2011-2022 走看看