Populating Next Right Pointers in Each Node--LeetCode

题目：

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：只是把每个节点都添加了一个next节点，对于一棵树，把next节点都进行赋值，而且是一个完全二叉树，可以使用层次遍历，为每一层都进行改造。

void Connect(BinTree* root){     if(root == NULL)         return ;     deque<BinTree*> de;     de.push_back(root);     int i,level=1;     BinTree* cur,*pre;      while(!de.empty())     {         cur = NULL;         pre = NULL;         for(i=0;i<level&&!de.empty();i++)         {             if(pre == NULL)             {               pre = de.front();               if(pre->left != NULL)                de.push_back(pre->left);               if(pre->right !=NULL)                de.push_back(pre->right);               de.pop_front();               cout<<pre->value<<endl;               continue;                 }                   if(cur == NULL)             {               cur = de.front();               if(cur->left != NULL)                de.push_back(cur->left);               if(cur->right !=NULL)                de.push_back(cur->right);               de.pop_front();               cout<<cur->value<<endl;               continue;              }             pre->next =cur;             pre = cur;             cur = de.front();             de.pop_front();             if(cur->left != NULL)               de.push_back(cur->left);             if(cur->right != NULL)               de.push_back(cur->right);               cout<<cur->value<<endl;                }                  if(cur == NULL)           pre->next == NULL;         else         {           pre->next = cur;           cur->next == NULL;           }         level *=2;                 }}

题目的要求是不用额外的空间，加入有一层已经弄好了，那么根据这一层是否可以初始化下一层？可以，从上一层头节点的左边开始即可。最后是改造后的层次遍历

  void helper(BinTree* pre_head){     if(pre_head->left == NULL )//|| pre_head->left== NULL)       return ;     BinTree* cur_first = pre_head->left;     BinTree* pre = pre_head;     BinTree* cur_second = pre_head->right;     while(pre != NULL&&pre->left != NULL)     {        cur_first->next = cur_second;        cur_first = cur_second;        pre = pre->next;        if(pre != NULL && pre->left != NULL)        {           cur_second = pre->left;           cur_first->next = cur_second;           cur_first = cur_second;           cur_second = pre->right;        }               }     if(cur_first != NULL)       cur_first->next =NULL;     helper(pre_head->left);}void Connect_second(BinTree* root){     if(root ==NULL)         return ;     root->next = NULL;     helper(root);}void order(BinTree* root){     BinTree* temp = root;     if(temp == NULL)      return;     while(temp != NULL)     {                cout<<temp->value<<endl;                temp = temp->next;     }     order(root->left);}

ps:这是一个满二叉树，所以对于每一层节点的个数都是一定的，假如题目中没有要求使用常量的空间，那么仍然可以使用层次遍历的方法来处理这个问题，对于每一层的头节点和每一层的尾节点都是知道的，既然要求使用常量的空间复杂度，那么就需要使用递归了！

对于一个函数，如果传递了当前层次的第一个节点pre，那么就可以处理下面那个层次，因为是一个满二叉树，pre节点横向是一个单链表，由这个单链表可以遍历下面层次的各个节点