117、Populating Next Right Pointers in Each Node II

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:

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

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TreeDepth-first Search

题意：是上一题的普适版，不要求满树！每层节点链接起来，类似B*树

thinking：

（1）上一题说过，，采用BFS 而不是DFS：

（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);}};

其实你已经错过了旅行的意义。