【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

两道题基本一样吧，只不过1限定节点一点会有左右孩子，2不一定。
但是思路是一样的。
递归。每次修改next指向的时候，顺便将左右孩子放在list中等待下一次循环调用。
如果list为空了，表明树已经循环结束。退出。

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);
    }
}