populating-next-right-pointers-in-each-node-ii

1、题目描述

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

2、思路

使用队列遍历每一层，然后每一层用NULL作为标志隔开每一层。

3、代码

方案1、根据2中的思路：

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)
			return;
		queue<TreeLinkNode *> q;
		q.push(root);
		q.push(NULL);   //每一层用一个NULL来判别
		TreeLinkNode *cur, *next;
		cur = q.front();
		q.pop();
		while (cur) {
			if (cur->left)
				q.push(cur->left);
			if (cur->right)
				q.push(cur->right);
			next = q.front();
			q.pop();
			if (next) {  //如果不为NULL，还未结束，则继续遍历这一层
				cur->next = next;
				cur = next;
			}
			else {  //遇到NULL，表示遍历上一层结束
				q.push(NULL);  //上一层结束是，下一层进队列也结束
				cur = q.front();
				q.pop();
			}
		}
	}
};

方案2、其它参考代码：

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) 
    {
        while(root)
        {
            TreeLinkNode dummy(-1), *prev;
            prev = &dummy;
            for(auto p = root; p; p = p->next)
            {
                if(p->left)
                {
                    prev->next = p->left;
                    prev = prev->next;
                }
                if(p->right)
                {
                    prev->next = p->right;
                    prev = prev->next;
                }
            }
            root = dummy.next; // 指向下一层的第一个节点
        }
    }
};