leetcode 116/117 Populating Next Right Pointers in Each Node 1/2

116 and 117 同一份代码： 无论什么二叉树，都可以。不一定是完全二叉树
 Given a binary tree

struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to 

NULL

.
Initially, all next pointers are set to 

NULL

.
Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

1
/  \
2    3
/ \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
/  \
2 -> 3 -> NULL
/ \  / \
4->5->6->7 -> NULL

使用一个层级遍历的思想。无论是不是

/**
* 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 || (!root->left && !root->right))
return;
deque<TreeLinkNode *> dp;
dp.push_back(root);

while (!dp.empty())
{
int size = dp.size();
TreeLinkNode *last = NULL;
for (int i = 0; i < size; i++)
{
TreeLinkNode *point = dp.back();
dp.pop_back();
if (point->right)
dp.push_front(point->right);
if (point->left)
dp.push_front(point->left);

point->next = last;
last = point;
}
}
}
};