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,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL

这个题目是LeetCode[Tree]: Populating Next Right Pointers in Each Node的后续，这里的迭代思路与之相同，只有一个地方的差别：由于树有可能是任意的二叉树，因此当获取下一个节点的时，需要用一个循环进行查找。

我的C++的代码实现如下：

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;
    }
};

LeetCode[Tree]: Populating Next Right Pointers in Each Node II

原文：http://blog.csdn.net/chfe007/article/details/41758063