树——populating-next-right-pointers-in-each-node（填充每个节点的next指针）

Posted 2020-10-03 一个不会coding的girl

问题：

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 toNULL.

Initially, all next pointers are set toNULL.

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

 

思路：

首先判断根节点是否为null，如果是，return；如果不是，继续。

设置节点node，用来指向每层的第一个节点；设置节点next，用来指向该层的每个节点。

当node的左孩子存在时，使用next来遍历该层的其它节点：

　　（1）让node左孩子的next指向node右孩子；

　　（2）若next.next不为null，让node右孩子的next指向next.next的左孩子。

 

代码：

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11         if(root==null)
12             return;
13         root.next = null;
14         TreeLinkNode node = root, next = node;
15         while(node.left!=null){
16             next = node;
17             while(next!=null){
18                 next.left.next = next.right;
19                 if(next.next!=null)
20                     next.right.next = next.next.left;
21                 next = next.next;
22             }
23             node = node.left;
24         }
25     }
26 }