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

题意：如同116题，对任意的二叉树找出其结点的next点。
思路：如同，116题，只是在递归时要先递归右树，然后再递归左树。

/**
* 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) {}
* };
*/
/**
* 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) {
if (root == NULL)
return;
if (root->left){
if (root->right)
root->left->next = root->right;
else{
TreeLinkNode *tmp = root->next;
while (tmp){
if (tmp->left || tmp->right){
if (tmp->left){
root->left->next = tmp->left;;
}
else{
root->left->next = tmp->right;
}
break;
}
else
tmp = tmp->next;
}
}
}

if (root->right){
TreeLinkNode *tmp = root->next;
while (tmp){
if (tmp->left || tmp->right){
if (tmp->left){
root->right->next = tmp->left;;
}
else{
root->right->next = tmp->right;
}
break;
}
else
tmp = tmp->next;
}
}
connect(root->right);
connect(root->left);
}
};

思路2：思路1的代码过于冗余，精简一下如下。
/**
* 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) {}
* };
*/
/**
* 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) {
if (root == NULL)
return;

TreeLinkNode *tmp = root->next;
while (tmp){
if (tmp->left || tmp->right){
break;
}
else
tmp = tmp->next;
}
TreeLinkNode *t = NULL;
if(tmp){
if (tmp->left)
t = tmp->left;
else
t = tmp->right;
}
if (root->right){
root->right->next = t;
}
if (root->left){
if (root->right)
root->left->next = root->right;
else{
root->left->next = t;
}
}
connect(root->right);
connect(root->left);
}
};