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Leetcode: Populating Next Right Pointers in Each Node II

时间:2014-09-18 07:30:53      阅读:306      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

在Populating Next Right Pointers in Each Node问题的基础上,难度20,方法一样。都是类似Binary Tree Level Order Traverse,都是把树看成一个无向图,然后用BFS的方式,需要记录每一层的ParentNumInQueue以及ChildNumInQueue, 初始值为1和0,以后每次ParentNumInQ减至0说明这一层已经遍历完毕,这一层的Child数将成为下一层的ParentNumInQ

 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         if (root == null) return;
12         LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
13         queue.add(root);
14         int ParentNumInQ = 1;
15         int ChildNumInQ = 0;
16         TreeLinkNode pre = null;
17         while (!queue.isEmpty()) {
18             TreeLinkNode cur = queue.poll();
19             ParentNumInQ--;
20             if (pre == null) {
21                 pre = cur;
22             }
23             else {
24                 pre.next = cur;
25                 pre = pre.next;
26             }
27             if (cur.left != null) {
28                 queue.add(cur.left);
29                 ChildNumInQ++;
30             }
31             if (cur.right != null) {
32                 queue.add(cur.right);
33                 ChildNumInQ++;
34             }
35             if (ParentNumInQ == 0) {
36                 ParentNumInQ = ChildNumInQ;
37                 ChildNumInQ = 0;
38                 pre.next = null;
39                 pre = null;
40             }
41         }
42     }
43 }

 

Leetcode: Populating Next Right Pointers in Each Node II

原文:http://www.cnblogs.com/EdwardLiu/p/3978460.html

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