leetcode: (41) Populating Next Right Pointers in Each Node

【Question】

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
/ \  / \

方法一：题目给出的是满二叉树，通过层序遍历，将pow(2,n)-1位置的结点的->next设NULL，其他的结点指向下一次将要遍历的结点

* 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) {
queue<TreeLinkNode*> que;
TreeLinkNode *p;
p=root;
if(p==NULL) return;
que.push(p);
int flag=1;
int count=0;
while(!que.empty())
{
count++;
p=que.front();
que.pop();
if(count==(pow(2,flag)-1)) {flag++;p->next=NULL;}
else p->next=que.front();
if (p->left==NULL) continue;
que.push(p->left);
que.push(p->right);
}
}
};

这种方法开辟了额外的空间。

方法二：

class Solution {
public:
void connect(TreeLinkNode *root) {
if(NULL == root) return;
TreeLinkNode* curLev;
while(root -> left != NULL){
curLev = root;
while(curLev != NULL){
curLev -> left -> next = curLev -> right;
if(curLev -> next != NULL)
curLev -> right -> next = curLev -> next -> left;
curLev = curLev -> next;
}
root = root -> left;
}
}
};