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.

Example:

Given the following binary tree,

After calling your function, the tree should look like:

 

 

这道题目和上一道不同的是，它不是完全二叉树。则不能通过节点数计算是否在哪一层

c++

class Solution {public:    void connect(TreeLinkNode *root) {                if(root==NULL) return;        queue
      
       
         > q;        TreeLinkNode* pre = NULL;        q.push(make_pair(1,root));        int y = 0;        while(!q.empty())        {            TreeLinkNode* temp = q.front().second;            int lever = q.front().first;            q.pop();            if(lever!=q.front().first||q.empty())            {                if(pre!=NULL)                    pre->next =temp;                temp->next = NULL;                pre = NULL;            }            else{                                if(lever!=y) {y=lever;pre = temp; pre->next =NULL;}                else { pre->next = temp;pre = temp;pre->next=NULL;                     }             }                     if(temp->left!=NULL) q.push(make_pair(lever+1,temp->left));            if(temp->right!=NULL) q.push(make_pair(lever+1,temp->right));                    }            }};