1.题目描述

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

2.解法分析

这道题目给了一个很强的约束，那就是可以假设树结果为满二叉树，满二叉树每一层的节点个数是确定的，如果将树中的元素按照层序遍历的方式编号，那么编号为n的节点在哪一层也是轻易可知的（假设编号从1开始，那么节点n所在层为log₂ⁿ⁺¹），同样，每层最后一个节点的编号也是已知的(m层的最后一个元素编号为2^m-1)，基于这个强约束，我决定用层序遍历的方式解答这个题目。

/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(root == NULL)return; queue<TreeLinkNode *> q; int count=0; int depth =1; TreeLinkNode * cur=NULL; q.push(root); while(!q.empty()) { cur=q.front(); q.pop(); count++; if(count == pow(2,depth)-1) { cur->next = NULL; depth++; } else { cur->next = q.front(); } if(cur->left!=NULL)q.push(cur->left); if(cur->right!=NULL)q.push(cur->right); } } };

代码一次通过，真爽！