116. Populating Next Right Pointers in Each Node

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 toNULL.
Initially, all next pointers are set toNULL.
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：此题相当于一个层次遍历，当每层遍历完后将next定为NULL，最开始的难题在于怎么设置链表的头节点，后来利用先初始化一个头节点head，当queue里面每出一个节点，将head的next定为该节点即可，最后上代码：

 void connect(TreeLinkNode *root) {        if(!root) return;        queue<TreeLinkNode*> queue1;        queue<TreeLinkNode*> queue2;        vector<queue<TreeLinkNode*> > v;        v.push_back(queue1);        v.push_back(queue2);          int cur=0;        int next=1;        v[cur].push(root);          while (!v[cur].empty()||!v[next].empty()) {              TreeLinkNode *head=new TreeLinkNode(-1);         while(!v[cur].empty()){        TreeLinkNode *t = v[cur].front();          v[cur].pop();          head->next=t;        head=head->next;        if (t->left != NULL)              v[next].push(t->left);          if (t->right != NULL)              v[next].push(t->right);      }      if(v[cur].empty()){        head->next=NULL;        cur=1-cur;        next=1-next;    }}    }