Populating Next Right Pointers in Each Node

题意：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 should be set to

NULL

.

Initially, all next pointers are set to

NULL

.

思路：看到这道题，很容易想到利用层序遍历的思想，得到每层节点的所有左右子节点，然后指向。

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;
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;
Q[rear]->next = Q[i]->right;
}
if (Q[i]->right != NULL)
{
Q[++rear] = Q[i]->right;
Q[rear]->next = (i+1<=index) ? Q[i+1]->left : NULL;
}
}
front = index + 1;
}
}
};