标题: | |
通过率: | 36.1% |
难度: | 中等 |
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 这个题目我感觉没必要一定是满二叉树,按层次遍历一遍。然后进行链表连接,需要注意的就是边界问题。开头和结尾的处理方法。 具体看代码
1 /** 2 * Definition for binary tree with next pointer. 3 * public class TreeLinkNode { 4 * int val; 5 * TreeLinkNode left, right, next; 6 * TreeLinkNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution {10 public void connect(TreeLinkNode root) {11 int count=1,tmp=0;12 LinkedListqueue=new LinkedList ();13 TreeLinkNode start=null,sec=null,pre=null;14 if(root!=null){15 queue.addLast(root);16 }17 while(!queue.isEmpty()){18 pre=queue.pollFirst();19 if(count==1)pre.next=null;20 if(pre.left!=null){21 queue.addLast(pre.left);22 tmp++;23 }24 if(pre.right!=null){25 queue.addLast(pre.right);26 tmp++;27 }28 29 for(int i=1;i