[leetcode] 117. Populating Next Right Pointers in Each Node II 解题报告

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题目链接：https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

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：可以利用一个虚头结点，记录要连接指针一层结点的第一个元素，然后一个指针从左到右依次连接上一层结点的左右结点。做这一题的时候我又回去看了一下我上一题，发现上一题要求用常量的辅助空间，我用了O(n)的空间。，。正好这个也可以解决那一题的问题。

代码如下：

/** * 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 *pHead = new TreeLinkNode(0), *pre = pHead;        while(root)        {            if(root->left)            {                pre->next = root->left;                pre = pre->next;            }            if(root->right)            {                pre->next = root->right;                pre = pre->next;            }            root = root->next;            if(!root)//如果到了一层的最后一个结点，就连接下一层的第一个结点            {                pre = pHead;                root = pHead->next;                pHead->next = NULL;            }        }    }};

参考：https://leetcode.com/discuss/67291/java-solution-with-constant-space

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