Leetcode 117. 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

s思路：
1. o(1)的空间，注定只能用iterative的方法了。参考https://discuss.leetcode.com/topic/1106/o-1-space-o-n-complexity-iterative-solution/6
2. 由于不规则的树结构，所以需要用pre，cur来找到新的连接关系的两端；还需要一个head表示每层的起点。三个指针的interplay在代码里面写得很清楚。每次把head赋给cur，然后根据cur->left、cur->right是否存在来更新连接:如果cur->left存在，又看pre是否已经存在：不存在则这个节点就是head,且pre=cur->left;存在则就把这个节点作为pre的next;对cur->right也同样判断。
3. 多体会！

//class Solution {public:    void connect(TreeLinkNode *root) {        //        TreeLinkNode* horizon=NULL,*vertical=root;        TreeLinkNode* head=root,*cur=NULL,*pre=NULL;        while(head){            cur=head;            pre=NULL;            head=NULL;            while(cur){                if(cur->left){                    if(!pre){                        head=cur->left;                        pre=head;                    }else{                        pre->next=cur->left;                            pre=pre->next;                    }                }                if(cur->right){                    if(!pre){                        head=cur->right;                        pre=head;                       }else{                        pre->next=cur->right;                        pre=pre->next;                      }                   }                cur=cur->next;                }        }    }};