LeetCode Populating Next Right Pointers in Each Node

题目

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

.

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 should look like:

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

输出每个节点右边的节点，bfs层序遍历即可。

代码：

/**
* 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) {
TreeLinkNode *pos,*next_level=root,*next_head=root;	//当前处理位置，下一层处理到位置，下一层头
while(next_head!=NULL)	//层序遍历
{
pos=next_head;
next_level=NULL;
next_head=NULL;
while(pos!=NULL)	//有未处理的层
{
if(next_level==NULL)	//处理左节点
{
next_level=pos->left;
next_head=next_level;
}
else
{
next_level->next=pos->left;
next_level=next_level->next;
}
if(pos->right==NULL)	//处理右节点
break;
next_level->next=pos->right;
next_level=next_level->next;
pos=pos->next;
}
}
}
};