117. Populating Next Right Pointers in Each Node II

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

不同于上一题，这题的二叉树并不是完全二叉树，我们不光需要提供first指针用来表示一层的第一个元素，同时也需要使用另一个lst指针表示该层上一次遍历的元素。那么我们只需要处理好如何设置last的next指针就可以了。
代码如下:

class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) {
return;
}

TreeLinkNode* p = root;
TreeLinkNode* first = NULL;
TreeLinkNode* last = NULL;

while(p) {
//设置下层第一个元素
if(!first) {
if(p->left) {
first = p->left;
} else if(p->right) {
first = p->right;
}
}

if(p->left) {
//如果有last，则设置last的next
if(last) {
last->next = p->left;
}
//last为left
last = p->left;
}

if(p->right) {
//如果有last，则设置last的next
if(last) {
last->next = p->right;
}
//last为right
last = p->right;
}

//如果有next，则转到next
if(p->next) {
p = p->next;
} else {
//转到下一层
p = first;
last = NULL;
first = NULL;
}
}
}
};

9.14

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这道题的关键之处就在于想清楚最后两层。！！！！！！