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

Follow up for problem "Populating Next Right Pointers in Each Node[/i]".What if the given tree could be any binary tree? Would your previous solution still work?Note:[/b]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:       }