二叉树横向连接

• 原来的树是这样的：

        1       /  \      2    3     / \  / \    4  5  6  7

• 将其转换成这样

        1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

• 节点结构如下：

  struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

• 解法如下：

class Solution {public:    void connect(TreeLinkNode *root) {        if(root == NULL) return ;        while(root->left){            TreeLinkNode * tmpRoot = root;                tmpRoot->left->next = tmpRoot->right;            while(tmpRoot->next){                tmpRoot->right->next = tmpRoot->next->left;                tmpRoot = tmpRoot->next;                if(tmpRoot->left)                    tmpRoot->left->next = tmpRoot->right;            }            root = root->left;        }    }};