[LeetCode] 076: 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

class Solution {
   
public:
	void connect(TreeLinkNode *root) {
   
		// Note: The Solution object is instantiated only once and is reused by each test case.
		// null node
		if(NULL == root)return;

		// initial
		queue<vector<TreeLinkNode*> > myQueue;
		vector<TreeLinkNode*> head;
		head.push_back(root);
		myQueue.push(head);

		// BFS
		while(!myQueue.empty()){
   
			vector<TreeLinkNode*> row = myQueue.front();
			myQueue.pop();

			// visit the first
			vector<TreeLinkNode*> next;
			TreeLinkNode* pre = row[0];
			if(pre->left != NULL){
   
				next.push_back(pre->left);
			}
			if(pre->right != NULL){
   
				next.push_back(pre->right);
			}

			// visit the remain
			for(int i = 1; i < row.size(); ++i){
   
				pre->next = row[i];
				if(row[i]->left != NULL){
   
					next.push_back(row[i]->left);
				}
				if(row[i]->right != NULL){
   
					next.push_back(row[i]->right);
				}
				pre = row[i];
			}

			// enqueue
			if(next.size() > 0)myQueue.push(next);
		}
	}
};