問題：

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.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

分析：

這道題叫我們對一課完全二叉樹進行操作，如果當前節點右邊有節點，就將next指標指向它，否則就置為空。我是用兩個list來完成的，一個用來儲存當前層的節點，另一個用來儲存下一層的節點（按順序），當前層處理完後，就將下一層的節點賦給當前層，一直到葉子節點。

程式碼：

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        //為空直接跳出
        if(root == NULL){
            return;
        }
        
        //當前層
        list<TreeLinkNode*> current;
        TreeLinkNode* cur_node;//當前節點
        TreeLinkNode* temp_node;//右邊的節點（沒有時為空）
        
        current.push_back(root);
        
        while( true ){
            list<TreeLinkNode*> temp;
            
            while( current.size() != 0 ){
                cur_node = current.front();
                current.pop_front();

                if(cur_node->left != NULL){
                    temp.push_back(cur_node->left);
                    temp.push_back(cur_node->right);
                }

                //當前節點右邊還有節點
                if(current.size()!=0){
                    temp_node = current.front();
                    cur_node->next = temp_node;
                }else{
                    cur_node->next = NULL;
                }
            }
           
            current = temp;
            
            if( current.size() == 0 && temp.size() == 0 ){
                break;
            }
        }
    }
};