117. Populating Next Right Pointers in Each Node II

Posted 2020-10-21 张乐乐章

 

 

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,

         1
       /        2    3
     / \        4   5    7

 

After calling your function, the tree should look like:

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

 1 public class Solution {
 2     public void connect(TreeLinkNode root) {
 3         TreeLinkNode head=null;
 4         TreeLinkNode prev =null;
 5         TreeLinkNode cur = root;
 6         
 7         while(cur!=null){
 8             //针对每一层，利用Prev进行next连接
 9             while(cur!=null){
10                 
11                 if(cur.left!=null){
12                     if(prev==null)//prev空，说明cur.left是下一层的头节点。
13                         head = cur.left;
14                     else//不为空，正常连接next指针，
15                         prev.next =cur.left;
16                     prev = cur.left; //移动prev
17                 }
18                 
19                 if(cur.right!=null){
20                  if(prev==null)//prev空，说明cur.right是下一层的头节点。
21                      head = cur.right;
22                 else //不为空，正常连接next指针，
23                      prev.next=cur.right;
24                 prev = cur.right;
25                 }
26                 
27                 cur = cur.next;
28             }
29             //此层结束，准备进入下一层。
30             cur = head;
31             head = null;
32             prev = null;
33         }
34     }
35 }