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题目：
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
代码：
/** * 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; deque<TreeLinkNode *> curr, next; curr.push_back(root); while ( !curr.empty() ) { TreeLinkNode dummy(-1); TreeLinkNode *pre = &dummy; while ( !curr.empty() ) { TreeLinkNode *tmp = curr.front(); curr.pop_front(); pre->next = tmp; if (tmp->left) next.push_back(tmp->left); if (tmp->right) next.push_back(tmp->right); pre = tmp; } pre->next = NULL; std::swap(curr, next); } } };
tips：
广搜思路（有些违规，因为不是const extra space）
每一层设立一个虚拟头结点，出队的同时pre->next = tmp
【Populating Next Right Pointers in Each Node II】cpp
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4505515.html
