【LeetCode】Populating Next Right Pointers in Each Node I & II

1、Populating Next Right Pointers in Each Node  Total Accepted: 12834 Total Submissions: 37308 My Submissions 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

2、Populating Next Right Pointers in Each Node II 
Total Accepted: 9074 Total Submissions: 30925 My Submissions
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

Discuss
두 문제는 기본적으로 똑같죠. 단지 1한정 노드에 한 번에 좌우 아이가 있을 뿐입니다. 2는 꼭 그렇지는 않습니다.하지만 사고방식은 같다.차례로 돌아가다.넥스트가 가리키는 것을 수정할 때마다, 왼쪽 아이를list에 놓고 다음 순환 호출을 기다립니다.목록이 비어 있으면 트리가 순환이 끝났음을 나타냅니다.물러나다.
Java AC

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
       if(root == null){
            return;
        }
        List<TreeLinkNode> list = new ArrayList<TreeLinkNode>();
        list.add(root);
        createNextNode(list);
    }
    public void createNextNode(List<TreeLinkNode> list){
        if (list.isEmpty() || list.size() == 0) {
			return;
		}
		int size = list.size();
		List<TreeLinkNode> tempList = new ArrayList<TreeLinkNode>();
		for (int i = 0; i < size; i++) {
			TreeLinkNode node = list.get(i);
			if (i == size - 1) {
				node.next = null;
			} else {
				node.next = list.get(i + 1);
			}
			if (node.left != null) {
				tempList.add(node.left);
			}
			if(node.right != null){
			    tempList.add(node.right);
			}
		}
		createNextNode(tempList);
    }
}