【populating-next-right-pointers-in-each-node-ii】

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

题意：把每一层节点链接起来

思路：层次遍历的思想，将一层的元素全部入队，然后将本层每个节点的子节点一次全部入队

class Solution{public:void connect(TreeLinkNode* root){if (root==NULL){return;}TreeLinkNode* tail = root;TreeLinkNode* tmp;queue<TreeLinkNode*> q;q.push(root);while (q.size()){tmp = q.front();q.pop();if (tmp->left!=NULL){q.push(tmp->left);}if (tmp->right!=NULL){q.push(tmp->right);}if(tmp==tail){tmp->next = NULL;tail = q.back();}else{tmp->next = q.front();}}}};