[LeetCode]Populating Next Right Pointers in Each Node II

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

思考：DFS会出错，层次遍历。

/**
 * 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;
        queue<TreeLinkNode *> s; //队列存储遍历结点
        s.push(root);
        while(!s.empty())
        {
            TreeLinkNode *p=s.front();
            s.pop();
            if(p->left)
            {
                s.push(p->left);
                if(p->right) p->left->next=p->right;
                else 
                {
                    TreeLinkNode *q=p->next;
                    while(q)
                    {
                        if(q->left) {p->left->next=q->left;break;}
                        else if(q->right) {p->left->next=q->right;break;}
                        else q=q->next;
                    }
                }
            }
            if(p->right)
            {
                s.push(p->right);
                TreeLinkNode *q=p->next;
                while(q)
                {
                    if(q->left) {p->right->next=q->left;break;}
                    else if(q->right) {p->right->next=q->right;break;}
                    else q=q->next;
                }
            }           
        }
    }
};