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;
		    }
	    }
    }
};