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 Node的代码一点都没改在这道题就ACCEPTED，我感觉好失败啊，所以，上一题应该有更简单的方法。

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);
		}
    }
};