199. Binary Tree Right Side View Java Solutions

Source: Internet
Author: User

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.

For example:
Given The following binary tree,

   1            <---/   2     3         <---\       5     4       <---

You should return [1, 3, 4] .

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      PublicList<integer>Rightsideview (TreeNode root) { Alist<integer> ans =NewArraylist<integer>(); -Addright (ans,0, root); -         returnans; the     } -      -      Public voidAddright (list<integer> ans,intHigh , TreeNode root) { -         if(Root = =NULL)return; +         if(ans.size () = =High ) Ans.add (root.val);//Add a single element to each layer -Addright (ans,high+1, root.right); +Addright (ans,high+1, root.left); A     } at}

The problem can be solved by recursion, each time the element is added from the far right, and each layer joins one. Left dial hand tree can join the situation is higher than the right sub-tree.

Such as:

1

/       \

2 5

/

4

In the case where element 4 is located. Although the code is recursive, the actual program execution time is O (n) complexity.

Solution 2:

The problem can also be used in sequence traversal, each layer of the value of the right node is added to the list.

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      PublicList<integer>Rightsideview (TreeNode root) { Alist<integer> ans =NewArraylist<integer>(); -         if(Root = =NULL)returnans; -deque<treenode> q =NewLinkedlist<treenode>(); the         intCurrent = 1,next = 0;//Use current and next to mark the number of elements in the next layer - TreeNode node; - q.addlast (root); -          while(Q.size () > 0){ +node =Q.removefirst (); -current--; +             if(Node.left! =NULL){ A Q.addlast (node.left); atnext++; -             } -             if(Node.right! =NULL){ - Q.addlast (node.right); -next++; -             } in             if(current = = 0){ - Ans.add (node.val);//Add the right element of the layer each time toCurrent =Next; +Next = 0; -             } the         } *         returnans; $     }Panax Notoginseng      -}

Actual run time:

Better solution one.

199. Binary Tree Right Side View Java solutions

Contact Us

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.