Google+ Followers

Saturday, December 29, 2012

[LeetCode] Populating Next Right Pointers in Each Node II 解题报告


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
» Solve this problem

[解题思路]
与上一题类似,唯一的不同是每次要先找到一个第一个有效的next链接节点,并且递归的时候要先处理右子树,再处理左子树。


[Code]
1:    void connect(TreeLinkNode *root) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      if(root== NULL) return;  
5:      TreeLinkNode* p = root->next;  
6:      while(p!=NULL)  
7:      {  
8:         if(p->left!=NULL)  
9:         {  
10:            p = p->left;  
11:            break;  
12:        }  
13:        if(p->right!=NULL)  
14:        {  
15:           p = p->right;  
16:           break;  
17:        }  
18:        p = p->next;  
19:      }  
20:      if(root->right!= NULL)  
21:      {        
22:        root->right->next = p;  
23:      }  
24:      if(root->left !=NULL)  
25:      {        
26:        root->left->next = root->right? root->right:p;        
27:      }  
28:      connect(root->right);  
29:      connect(root->left);      
30:    }  


[Note]
1. Line 6, while loop, not if
For example,
 Level 1                      1
                            /              \
 Level 2              2               3
                     /        \                \
 Level 3       4         5                 6
                 /
 Level 4   7

When processing Level 3, while loop is necessary for finding a valid next.


Update 03/09/2014 Add a non-recursion version

1:       void connect(TreeLinkNode *root) {  
2:            TreeLinkNode* cur = root, *next = NULL;  
3:            while(cur!=NULL)  
4:            {  
5:                 TreeLinkNode *p = cur, *k= NULL;  
6:                 while(p!=NULL)  
7:                 {  
8:                      TreeLinkNode* sub = getLinkedLeftNode(p);  
9:                      if(sub != NULL)  
10:                      {  
11:                           if(next == NULL)  
12:                           {  
13:                                next = sub;  
14:                                k = sub;  
15:                           }  
16:                           else  
17:                                k->next = sub;  
18:                           while(k->next !=NULL) // ietrate to the tail  
19:                                k = k->next;  
20:                      }  
21:                      p = p->next;  
22:                 }  
23:                 cur = next;  
24:                 next = NULL;  
25:            }  
26:       }  
27:       TreeLinkNode* getLinkedLeftNode(TreeLinkNode * root)  
28:       {  
29:            if(root->left != NULL && root->right != NULL)  
30:                 root->left->next = root->right;  
31:            if(root->left != NULL)  
32:                 return root->left;  
33:            if(root->right != NULL)  
34:                 return root->right;  
35:            return NULL;  
36:       }  


Post a Comment