[LeetCode] 117. Populating Next Right Pointers in Each Node II, leetcodepointers

Source: Internet
Author: User

[LeetCode] 117. Populating Next Right Pointers in Each Node II, leetcodepointers
[Question]

Follow up for problem "Populating Next Right Pointers in Each Node ".

What if the given tree cocould be any binary tree? Wocould 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 shoshould look like:

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

[Analysis]

None

[Code]

`/********************************** Date: * Author: SJF0115 * Subject: 117. populating Next Right Pointers in Each Node II * Source: Pull result: AC * Source: LeetCode * conclusion: * *********************************/# include <iostream> # include <queue> using namespace std; struct TreeLinkNode {int val; TreeLinkNode * left; TreeLinkNode * right; TreeLinkNode * Next; TreeLinkNode (int x): val (x), left (NULL), right (NULL), next (NULL) {}}; class Solution {public: void connect (TreeLinkNode * root) {if (root = NULL) {return;} // if queue <TreeLinkNode *> cur; queue <TreeLinkNode *> next; cur. push (root); TreeLinkNode * p, * pre; while (! Cur. empty () {pre = NULL; // The current layer traverses while (! Cur. empty () {// out queue p = cur. front (); cur. pop (); // horizontally connected if (pre! = NULL) {pre-> next = p;} // if pre = p; // next stores the next layer of nodes. // The left subtree is not empty and is added to the queue. if (p-> left) {next. push (p-> left);} // if // Add the right subtree to the queue if (p-> right) {next. push (p-> right) ;}// if} // while p-> next = NULL; swap (next, cur) ;}// while }}; // create a binary tree int CreateBTree (TreeLinkNode * & T) {int data; // input the value of the node in the binary tree in the first order, -1 indicates the empty tree cin> data; if (data =-1) {T = NULL;} else {T = new TreeLinkNode (data ); // construct the left subtree CreateBTree (T-> left); // construct the right subtree CreateBTree (T-> right);} return 0 ;} // output void LevelOrder (TreeLinkNode * root) {if (root = NULL) {return;} // if TreeLinkNode * p = root, * q; while (p) {q = p; // horizontal output while (q) {cout <q-> val <"->"; q = q-> next ;} // while if (q = NULL) {cout <"NULL" <endl;} // if p = p-> left ;} // while} int main () {Solution solution; TreeLinkNode * root (0); CreateBTree (root); solution. connect (root); LevelOrder (root );}`

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• Sales Support

1 on 1 presale consultation

• After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.