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:
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
解题思路：和前面那题比较像，只是现在的树形是普通的树形。用指针模拟队列。最巧妙的应该是用head指针指向下一层的开头。
/** * 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) { if(!root)return ; TreeLinkNode* head=root;//The left most node in the lower level TreeLinkNode* pre=NULL;//The previous node in the lower level TreeLinkNode* cur=NULL;//The current node in the upper level while(head){ cur=head; pre=NULL; head=NULL; while(cur){ if(cur->left){ if(!pre)head=cur->left; else pre->next=cur->left; pre=cur->left; } if(cur->right){ if(!pre)head=cur->right; else pre->next=cur->right; pre=cur->right; } cur=cur->next; } } } };
117. Populating Next Right Pointers in Each Node II
原文：http://www.cnblogs.com/tsunami-lj/p/6476780.html