leetcode Populating Next Right Pointers in Each Node(*)

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

考察观察变通能力，举例如下：对于1来说，1->left->next=1->right；此时，我们可以立足于第二层2->next=3来遍历第二层的链表，使得4->5->6->7，然后立足第三层向下遍历；

/**
* 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* temp=root;
if(temp->left)temp->left->next=temp->right;
while(temp->next&&temp->left){
temp->right->next=temp->next->left;
temp=temp->next;
temp->left->next=temp->right;
}
connect(root->left);
}
};