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                                                                                                                          本文地址

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

分析：直接考虑普通的二叉树：层序遍历二叉树，把每一层中前一个节点的next指向后一个节点，使用队列辅助层序遍历时，在队列中用NULL来分割每层的节点，能够经过两题测试的代码以下：

 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做为队列中每层节点之间的间隔
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 };

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