leetcode_117_Populating Next Right Pointers in Each Node II

描述：

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

思路：

就是用两个list存储相邻两层的结点，把第i+1层结点存储到list中的同时，分别将第i层的各个节点指向下一个，跟116题Populating Next Right Pointers in Each Node I一个思路，就不重复写了。

代码：

/**
 * 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<TreeLinkNode>temp=new ArrayList<TreeLinkNode>();
	      TreeLinkNode node=null,tempNode=null;
	      list.add(root);
	      while(true)
	      {
	    	  tempNode=list.get(0);
	    	  if(tempNode.left!=null)
	    		  temp.add(tempNode.left);
	    	  if(tempNode.right!=null)
	    		  temp.add(tempNode.right);
	    	  for(int i=1;i<list.size();i++)
	    	  {
	    		  node=list.get(i);
	    		  if(node.left!=null)
		    		  temp.add(node.left);
		    	  if(node.right!=null)
		    		  temp.add(node.right);
		    	  tempNode.next=node;
		    	  tempNode=node;
	    	  }
	    	  if(temp.size()!=0)
	    	  {
	    		 list=temp;
	    		 temp=new ArrayList<TreeLinkNode>();
	    	  }
	    	  else
	    		  break;
	      }
	      return;
	 }
}

结果：