problem：
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
Tree Depth-first
Searchthinking：
（1）上一题说过，采用BFS 而不是DFS：http://blog.csdn.net/hustyangju/article/details/45242365
（2）采用queue结构存储每层的结点
code：
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root==NULL)
return;
queue<TreeLinkNode*> queue0;
queue0.push(root);
level_visit(queue0);
return;
}
protected:
void level_visit(queue<TreeLinkNode*> queue1)
{
if(queue1.empty())
return;
queue<TreeLinkNode*> queue2=queue1;
queue<TreeLinkNode*> queue3;
TreeLinkNode *tmp=queue1.front();
queue1.pop();
tmp->next=NULL;
while(!queue1.empty())
{
TreeLinkNode *tmp2=queue1.front();
queue1.pop();
tmp2->next=NULL;
tmp->next=tmp2;
tmp=tmp2;
}
while(!queue2.empty())
{
TreeLinkNode *node=queue2.front();
queue2.pop();
if(node->left!=NULL)
queue3.push(node->left);
if(node->right!=NULL)
queue3.push(node->right);
}
level_visit(queue3);
}
};leetcode || 117、Populating Next Right Pointers in Each Node II
原文地址：http://blog.csdn.net/hustyangju/article/details/45242577
