LeetCode_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

　　方法一： constant extra space

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
      if( !root  || ( !root->left && !root->right)) return ;
        
        root->left->next = root-> right;
        if(root->next)
            root->right->next = root->next->left;    
        connect(root->left);
        connect(root->right);
        
    }
};

 方法二： 空间使用上貌似不符合要求

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
      if(NULL == root) return ;
      queue<TreeLinkNode *> myqueue, tpqueue;
      TreeLinkNode *p,*pre;
      myqueue.push(root);
      while(!myqueue.empty()){
        pre = myqueue.front();myqueue.pop();
        if(pre->left) tpqueue.push(pre->left);
        if(pre->right) tpqueue.push(pre->right);
        while(!myqueue.empty()){
            p = myqueue.front();myqueue.pop();
            if(p->left) tpqueue.push(p->left);
            if(p->right) tpqueue.push(p->right);
            pre->next = p;pre = p;
        }
        
        myqueue.swap(tpqueue);
      }
    }
};

 方法三： 方法一的非递归实现，符合题目的空间要求

class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
      if(NULL == root) return;
      TreeLinkNode *head = root;
      TreeLinkNode *tp;
      while(head->left){
        tp = head->left;
        while(head){
           if(head->left)
                head->left->next = head ->right;
            if(head ->right && head->next)
                head->right->next = head->next->left;
            head = head ->next;    
        }
        head = tp;
      }
    }
};