LeetCode[Tree]: 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,

After calling your function, the tree should look like:

我的C++代码实现如下：

class Solution {public:    void connect(TreeLinkNode *root) {        for (TreeLinkNode *levelFirstNode = root; levelFirstNode != nullptr; levelFirstNode = getNext(levelFirstNode)) {            for (TreeLinkNode *curNode = levelFirstNode; curNode != nullptr; curNode = curNode->next) {                if (curNode->left) curNode->left->next = curNode->right ? curNode->right : getNext(curNode->next);                if (curNode->right) curNode->right->next = getNext(curNode->next);            }        }    }private:    TreeLinkNode *getNext(TreeLinkNode *node) {        while (node) {            if (node->left) return node->left;            if (node->right) return node->right;            node = node->next;        }        return nullptr;    }};

其中，getNext函数获取的是从当前节点开始的下一层的第一个节点。但是这种算法的时间性能表现非常不好（如下图所示），希望发掘更好更快的算法来解决这个问题。