leetcode No117. Populating Next Right Pointers in Each Node II

Question

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

和leetcode No116. Populating Next Right Pointers in Each Node不同，二叉树不再是完全二叉树

Algorithm

leetcode No116. Populating Next Right Pointers in Each Node解法：
http://blog.csdn.net/u011391629/article/details/52415154

这题有两种解法：
解法一：是延续leetcode No116. Populating Next Right Pointers in Each Node的思路，根据上一层的next信息求解下一层，可以申请一个节点作为这一层的头结点，这样方面知道下一层的头结点

解法二：是二叉树按层遍历，pre结点保存上一个结点信息，然后把结点接起来。

Accepted Code

解法一：

class Solution {public:    void connect(TreeLinkNode *root) {        if(root==NULL)            return;        while(root!=NULL){            TreeLinkNode* nextLevel=new TreeLinkNode(-1);  //for record next level Node            TreeLinkNode* cur=nextLevel;            while(root!=NULL){                if(root->left!=NULL){                    cur->next=root->left;                    cur=cur->next;                }                if(root->right!=NULL){                    cur->next=root->right;                    cur=cur->next;                }                root=root->next;            }            root=nextLevel->next;        }    }};

解法二：

class Solution {public:    void connect(TreeLinkNode *root) {        if(root==NULL)            return;        while(root!=NULL){            TreeLinkNode* nextLevel=new TreeLinkNode(-1);  //for record next level Node            TreeLinkNode* cur=nextLevel;            while(root!=NULL){                if(root->left!=NULL){                    cur->next=root->left;                    cur=cur->next;                }                if(root->right!=NULL){                    cur->next=root->right;                    cur=cur->next;                }                root=root->next;            }            root=nextLevel->next;        }    }};