Populating Next Right Pointers in Each Node II - LeetCode 117

题目描述：
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:

You may only use constant extra space.
For example,
Given the following binary tree,
         1
       /  \
      2    3
     / \    \
    4   5    7
After calling your function, the tree should look like:
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL
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分析：此题的思路和Populating Next Right Pointers in Each Node（http://blog.csdn.net/bu_min/article/details/45921295）是一样的，只是处理的时候，要跳过二叉树的空节点。next指针要指向其右边第一个非空节点。
只需在上一题的基础上跳过二叉树中的空节点，处理每一层的时候，注意判断其左右孩子是否为空，只将非空节点放入向量和队列中。

以下是C++实现代码：

/*/////////////////////////44ms////////*//** * 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) {        if(root == NULL)            return;        TreeLinkNode *cur = root;        queue<TreeLinkNode* > q;    root->next = NULL; //修改root的next指针        if(cur != NULL)    {        q.push(cur);        while(!q.empty()) //此处判断是否最后一层处理完        {        int len = 0;        vector<TreeLinkNode*> vec; //存放即将要处理得层的每个节点的左右节点        while(!q.empty()) //此处判断某一层是否处理完        {            TreeLinkNode *cur = q.front();            q.pop();            if(cur->left != NULL)  //依次将非空的左右孩子放入vec中            {                vec.push_back(cur->left);                  len++;            }            if(cur->right != NULL)            {                vec.push_back(cur->right);                len++;            }        }        if(len > 0) 当有非空左右孩子时，才修改next指针，将其指向右边的第一个非空节点        {            for(int i = 0; i < len-1; i++)             {            q.push(vec[i]);            vec[i]->next = vec[i+1];            }            q.push(vec[len-1]);            vec[len - 1]->next = NULL;        }        }        }        return;    }};