116. Populating Next Right Pointers in Each Node

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.

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

默认每个节点的next 都是空。
2层循环，第一层循环是层数，用level_start控制，
第二层循环，循环每层的每个节点，来构建next 指针，
构建next指针时，有3种情况：1 左指针指向右指针
2 右指针不是当层最末尾的指针，右指针指向父亲节点next的左指针。
3
右指针是当层最末尾的指针，不用处理，因为默认每个节点的next都是指向null的。

1 public class Solution {
2     public void connect(TreeLinkNode root) {
3         TreeLinkNode level_start=root;
4         while(level_start!=null){
5             TreeLinkNode cur=level_start;
6             while(cur!=null){
7                 if(cur.left!=null) cur.left.next=cur.right;
8                 if(cur.right!=null && cur.next!=null) cur.right.next=cur.next.left;
9
10                 cur=cur.next;
11             }
12             level_start=level_start.left;
13         }
14     }
15 }