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

思路：BFS

class Solution {
public:
void connect(TreeLinkNode *root) {
if (!root) return;
queue<TreeLinkNode*> que;
que.push(root);
int cur = 1;
int nex = 0;
while (!que.empty()){
while (cur--){
TreeLinkNode* p = que.front();
que.pop();

if (p->left){
que.push(p->left);
nex++;
}
if (p->right){
que.push(p->right);
nex++;
}
if (cur == 0) //当前层的最后一个节点
p->next = nullptr;
else
p->next = que.front();
}
cur = nex;
nex = 0;
}
}
};