Populating Next Right Pointers in Each Node II

Description:

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

Code:

 1  void connect(TreeLinkNode *root) {
 2         if (!root)
 3             return ;
 4         deque<TreeLinkNode*>a;
 5         deque<TreeLinkNode*>b;
 6         a.push_back(root);
 7         
 8         while (!a.empty())
 9         {
10             while (!a.empty())
11             {
12                 TreeLinkNode* p = a.front();
13                 a.pop_front();
14                 if(p->left)
15                     b.push_back(p->left);
16                 if(p->right)
17                     b.push_back(p->right);
18                
19                if (a.empty())  
20                     p->next = NULL; 
21                else
22                     p->next = a.front();
23             }
24             a = b;
25             b.clear();
26         }
27     }

 

posted @ 2015-06-20 15:49  Rosanne  阅读(196)  评论(0)  编辑  收藏  举报