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

思路：
对于root，如果存在左右节点则，左节点的next是右节点。如果只存在之一的话，那么那么的next节点是root的next中最左边的节点。注意，因为子节点要用到父节点的next信息，所以先遍历右节点后遍历左节点。
代码：

    void connect(TreeLinkNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(root){
            if(root->left){
                if(root->right){
                    root->left->next = root->right;
                }
                else{
                    TreeLinkNode *tmp = root->next;
                    while(tmp && !root->left->next){
                        if(tmp->left)
                            root->left->next = tmp->left;
                        else
                            root->left->next = tmp->right;
                        tmp = tmp->next;
                    }
                }
            }
            if(root->right){
                TreeLinkNode *tmp = root->next;
                while(tmp && !root->right->next){
                    if(tmp->left)
                        root->right->next = tmp->left;
                    else
                        root->right->next = tmp->right;
                    tmp = tmp->next;
                }
            }
            connect(root->right);
            connect(root->left);
        }
    }