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：直接考慮普通的二叉樹：層序遍歷二叉樹，把每一層中前一個節點的next指向後一個節點，使用隊列輔助層序遍歷時，在隊列中用NULL來分割每層的節點，能夠經過兩題測試的代碼以下：

 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做爲隊列中每層節點之間的間隔
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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