leetcode-116. 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

思路：起初觉得是个很难的题，但是后来发现了Note里面的限制（给定二叉树是完全二叉树），所以简单好多，只需要逐层处理即可。

Accepted Code：

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         if(root==nullptr)
13         return;
14         TreeLinkNode* pre=root;
15         TreeLinkNode* cur=nullptr;
16         while(pre->left)
17         {
18             cur=pre;
19             while(cur)
20             {
21                 cur->left->next=cur->right;
22                 if(cur->next)
23                 {
24                     cur->right->next=cur->next->left;
25                     cur=cur->next;
26                 }else
27                 break;
28             }
29             pre=pre->left;
30         }
31     }
32 };