LeetCode:Populating Next Right Pointers in Each Node I II

LeetCode:Populating Next Right Pointers in Each Node 
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

 
 
LeetCode:Populating Next Right Pointers in Each Node II                                                                                                                            본문 주소
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

분석: 일반적인 두 갈래 트리를 직접 고려한다. 층층이 두 갈래 트리를 훑어보고 각 층의 이전 노드의next를 다음 노드를 가리키며 대기열을 보조하여 층층이 훑어볼 때 대기열에서 NULL로 각 층의 노드를 분할한다. 두 문제를 통해 테스트할 수 있는 코드는 다음과 같다.

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         // IMPORTANT: Please reset any member data you declared, as
13         // the same Solution instance will be reused for each test case.
14         if(root == NULL)return;
15         queue<TreeLinkNode*> myqueue;
16         myqueue.push(root);
17         myqueue.push(NULL);//NULL 
18         TreeLinkNode *pre = NULL;
19         while(myqueue.empty() == false)
20         {
21             TreeLinkNode *p = myqueue.front();
22             myqueue.pop();
23             if(p != NULL)
24             {
25                 if(p->left)myqueue.push(p->left);
26                 if(p->right)myqueue.push(p->right);
27             }
28             else if(myqueue.empty() == false)
29                 myqueue.push(NULL);
30             if(pre != NULL)pre->next = p;
31             pre = p;
32         }
33     }
34 };

[판권 성명] 전재 출처를 밝혀 주십시오:http://www.cnblogs.com/TenosDoIt/p/3437497.html