首页 > 其他 > 详细

Populating Next Right Pointers in Each Node II

时间:2015-01-04 13:35:52      阅读:262      评论:0      收藏:0      [点我收藏+]

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
与Populating Next Right Pointers in Each Node 基本一致  只是在向队列中加入子节点时要对左右节点分别判断 代码如下:
public class Solution {
    public void connect(TreeLinkNode root) {
        if(root==null) return ;
		int count=1;
		int level=0;
		Queue<TreeLinkNode> que =new LinkedList<TreeLinkNode>();
		que.offer(root);
		while(que.isEmpty()!=true){
			level=0;
			for(int i=0;i<count;i++){
				root=que.peek();
				que.poll();
				if(i<count-1){
				    root.next=que.peek();
				}else{
				    root.next=null;
				}
				if(root.left!=null){
					que.offer(root.left);
					level++;
				}
				if(root.right!=null){
				    que.offer(root.right);
					level++;
				}
			}
			count=level;
		}
    }
}


Populating Next Right Pointers in Each Node II

原文:http://blog.csdn.net/u012734829/article/details/42389677

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!