| 标题： | Populating Next Right Pointers in Each Node |
| 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 LinkedList<TreeLinkNode> queue=new LinkedList<TreeLinkNode>(); 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<count;i++){ 30 start=queue.pollFirst(); 31 pre.next=start; 32 pre=start; 33 if(start.left!=null){ 34 queue.addLast(start.left); 35 tmp++; 36 } 37 if(start.right!=null){ 38 queue.addLast(start.right); 39 tmp++; 40 } 41 } 42 count=tmp; 43 tmp=0; 44 if(start!=null) 45 start.next=null; 46 } 47 48 } 49 }