标题: | Populating Next Right Pointers in Each Node II |
通过率: | 31.7% |
难度: | 难 |
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
本题跟第一个版本一样,就是按层次遍历就行了。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 }