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

最新推荐文章于 2024-01-17 15:12:49 发布

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本文详细介绍了如何使用常数额外空间解决任意二叉树中节点的next指针问题，包括解题思路、代码实现以及非递归版本。通过遍历树并寻找有效的next链接节点，实现树节点间的连接。

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,

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:       }