leetcode_question_117 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

 

leetcode_question_116 Populating Next Right Pointers in Each Node의 코드를 이 문제에서 하나도 고치지 않고 ACCEPTED를 받았습니다. 저는 실패했다고 느꼈습니다. 그래서 이전 문제는 더 간단한 방법이 있을 것입니다.

BFS:

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root==NULL)return;
    	vector<TreeLinkNode*> vec;
		vec.push_back(root);
		int count = 1;
		while(!vec.empty())
		{
			if(count > 1) vec[0]->next = vec[1];
			else vec[0]->next = NULL;
			if(vec[0]->left) vec.push_back(vec[0]->left);
			if(vec[0]->right) vec.push_back(vec[0]->right);
			vec.erase(vec.begin());
			count--;

			if(count == 0)
				count = vec.size();
		}
    }
};

Recursive:

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    TreeLinkNode* rightNeighbor(TreeLinkNode* left)
{
	if(left->next == NULL) return NULL;
	else if(left->next->left)
		return left->next->left;
	else if(left->next->right)
		return left->next->right;
	else return rightNeighbor(left->next);
}

void connectHelper(TreeLinkNode* root)
{
	if(root)
	{
		if(root->right)
			root->right->next = rightNeighbor(root);
		if(root->left)
		{
			if(root->right)
				root->left->next = root->right;
			else
				root->left->next = rightNeighbor(root);
		}
		connectHelper(root->right);
		connectHelper(root->left);
	}
}

void connect(TreeLinkNode *root) {
        if(root)
		{
			root->next = NULL;
			if(root->right)
				root->right->next = NULL;
			if(root->left)
			{
				if(root->right) 
					root->left->next = root->right;
				else
					root->left->next = NULL;
			}
			connectHelper(root->right);
			connectHelper(root->left);
		}
    }
};