Populating Next Right Pointers in Each Node II

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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

思路：在前面I的基础上进行一些改进。下层节点的next指针指向需要对上一层的所有节点的左右节点进行遍历，如果不为NULL，则指向。例如节点5的next指向，需要遍历3节点的左右子节点，找到7，则指向。

class Solution {public:    void connect(TreeLinkNode *root) {        /*****************************************         * 队列的长度最好取10000以上，前面WA是因为队列的长度太小了         * ****************************************/        TreeLinkNode* Q[10000];        int front = 0, rear = -1;        int i,j;        Q[++rear] = root;          if (Q[rear] == NULL)        {            return;        }        int index;        while(front <= rear)        {            index = rear;            for(i = front; i <= index; i++)            {                if (Q[i]->left != NULL)                {                    Q[++rear] = Q[i]->left;                      if (Q[i]->right != NULL)                    {                        Q[rear]->next = Q[i]->right;                    }                         else                     {                        for(j = i+1; j<=index; j++)                        {                            if (Q[j]->left != NULL)                            {                                Q[rear]->next = Q[j]->left;                                break;                            }                                if (Q[j]->right != NULL)                            {                                Q[rear]->next = Q[j]->right;                                break;                            }                            }                        }                    }                if (Q[i]->right != NULL)                {                    Q[++rear] = Q[i]->right;                    for(j = i+1; j<=index; j++)                        {                            if (Q[j]->left != NULL)                            {                                Q[rear]->next = Q[j]->left;                                break;                            }                                if (Q[j]->right != NULL)                            {                                Q[rear]->next = Q[j]->right;                                break;                            }                            }                 }            }            front = index + 1;        }    }};

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