Leetcode 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

题意和116一样，只不过这里的树不保证每一个节点都有左右孩子。
找next的时候一定要找到有孩子的next才行，而且为了避免next的next没有连接的情况，应该先遍历右子树。

/**
* 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) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) return ;
if(root->right)
{
TreeLinkNode *p=root;
while(p->next)
{
p=p->next;
if(p->left)
{
root->right->next=p->left;
break;
}
else if(p->right)
{
root->right->next=p->right;
break;
}
}connect(root->right); //先遍历右子树
}
if(root->left)
{
if(root->right)
root->left->next=root->right;
else
{
TreeLinkNode *p=root;
while(p->next)
{
p=p->next;
if(p->left)
{
root->left->next=p->left;
break;
}
else if(p->right)
{
root->left->next=p->right;
break;
}
}
}
connect(root->left);
}
}
};