Populating Next Right Pointers in Each Node(leetcode)

题目：

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

题目来源：这两个题目是一样的

https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

解题思路：层次遍历

#include<iostream>#include<queue>using namespace std;struct TreeLinkNode {int val;TreeLinkNode *left, *right, *next;TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}};void connect(TreeLinkNode *root) {if(root==NULL)return ;queue<TreeLinkNode *> q;q.push(root);int k=0,level=1;while(!q.empty()){k=level;level=0;for(int i=0;i<k;i++){TreeLinkNode *temp=q.front();q.pop();if(temp->left!=NULL){q.push(temp->left);level++;}if(temp->right!=NULL){q.push(temp->right);level++;}if(i!=k-1)temp->next=q.front();elsetemp->next=NULL;}}}int main(){TreeLinkNode *root=new TreeLinkNode(1);root->left=new TreeLinkNode(2);root->right=new TreeLinkNode(3);root->left->left=new TreeLinkNode(4);root->left->right=new TreeLinkNode(5);root->right->right=new TreeLinkNode(6);system("pause");return 0;}