LeetCode Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-09-24 12:05:40 发布

转载最新推荐文章于 2021-09-24 12:05:40 发布 · 43 阅读

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原文链接：http://www.cnblogs.com/birdhack/p/4158823.html

本文介绍了一种在任意二叉树中填充每个节点的Next指针的方法，以便相同层级的相邻节点能够通过Next指针相连。该方法只使用常数额外空间，并详细解释了针对左节点和右节点的不同处理逻辑。

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Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

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

因此，对一个节点需要向右找到第一个节点。

对于left，如果right不存在，就在father的next节点去找left/right，依次找下去。

对于right，直接在father的next节点开始找。

 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     TreeLinkNode node;
11     public void connect(TreeLinkNode root) {
12         if (root==null) {
13             return;
14         }
15 
16         TreeLinkNode left = root.left;
17         TreeLinkNode right = root.right;
18 
19         if (left != null) {
20             left.next = right;
21             if (right == null) {
22                 node = root.next;
23                 while (node != null && left.next == null) {
24                     left.next = node.left;
25                     if (left.next == null) {
26                         left.next = node.right;
27                     }
28                     node = node.next;
29                 }
30             }
31         }
32 
33         if (right != null) {
34             node = root.next;
35             while (node != null && right.next == null) {
36                 right.next = node.left;
37                 if (right.next == null) {
38                     right.next = node.right;
39                 }
40                 node = node.next;
41             }
42         }
43 
44         connect(right);
45         connect(left);
46     }
47 }