[Leetcode] populating next right pointers in each node

Problem description:

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 shoshould be setNULL.

Initially, all next pointers are setNULL.

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 shoshould look like:

         1 -> NULL       /        2 -> 3 -> NULL     / \  /     4->5->6->7 -> NULL

Basic Ideas:

Use a queue to save the nodes at the same layer that need to be linked.

Code:

void connect(TreeLinkNode *root) {                 if(root == NULL)            return;        queue<TreeLinkNode*> _queue;        _queue.push(root);                 TreeLinkNode head(0);        TreeLinkNode* pre = &head;                 queue<TreeLinkNode*> second;        while(!_queue.empty())        {                         TreeLinkNode* tmp = _queue.front();            _queue.pop();            if(tmp->left != NULL)                second.push(tmp->left);            if(tmp->right != NULL)                second.push(tmp->right);            if(_queue.empty())            {                pre->next = tmp;                tmp->next = NULL;                pre = &head;                                 while(!second.empty())                {                    _queue.push(second.front());                    second.pop();                                     }            }            else              {                pre->next = tmp;                pre = tmp;            }        }        // return root;    }

[Leetcode] populating next right pointers in each node