Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; } Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,1 / \ 2 3 / \ / \ 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ / \ 4->5->6->7 -> NULL
思路：一般层次遍历需要申请队列，但是有了next指针，我们只需记录每层最左的节点，所以可以做到use constant extra space
class Solution {public: void connect (TreeLinkNode *root){ TreeLinkNode* parent; TreeLinkNode* current; TreeLinkNode* nextParent = root; while(nextParent){ //new level started parent = nextParent; nextParent = parent->left; current = nextParent; if(!current) return; //add next pointer current->next = parent->right; current = current->next; if(!current) return; while(parent->next){ parent = parent->next; current->next = parent->left; current = current->next; if(!current) return; current->next = parent->right; current = current->next; if(!current) return; } } }};