# Leetcode 199:binary Tree Right Side View

Source: Internet
Author: User

`1 //My Code2  PackageLeetcode;3 /**4 * 199. Binary Tree right Side View5 * Address:https://leetcode.com/problems/binary-tree-right-side-view/6 * Given A binary tree, imagine yourself standing on the right side of it,7 * Return the values of the nodes you can see ordered from top to bottom.8  * 9  */Ten ImportJava.util.*; One  A  Public classBinarytreerightsideview { -      Public Static voidMain (string[] args) { -Treenode[] Tree =NewTreenode[9]; the          for(inti = 1; I < 9; i++){ -Tree[i] =NewTreeNode (i); -         } -Treenode.link (tree, 1, 2, 3); +Treenode.link (Tree, 2, 4, 5); -Treenode.link (Tree, 3, 6, 1); +Treenode.link (Tree, 5, 7, 8); A         //First Order Traversal atTreenode.preprint (tree[1]); -         //Solution -list<integer> result = Rightsideview (tree[1]); - System.out.println (result); -          -         //Solution2 inlist<integer> result2 = RightSideView2 (tree[1]); - System.out.println (RESULT2); to          +     } -     /** the * Method One: Time complexity is high, need O (n), n is the number of nodes in the tree *      * @paramRoot \$      * @returnPanax Notoginseng      */ -      Public StaticList<integer>Rightsideview (TreeNode root) { thelist<integer> result =NewArraylist<>(); +queue<treenode> queue =NewLinkedlist<>(); A         intChildnum = 0; the             if(Root! =NULL){ + Queue.offer (root); -                  while(!Queue.isempty ()) { \$TreeNode node =Queue.poll (); \$                     if(node.left!=NULL){ - Queue.offer (node.left); -childnum++; the                     } -                     if(Node.right! =NULL)Wuyi                     { the Queue.offer (node.right); -childnum++; Wu                     } -System.out.println ("queue.size () =" +queue.size ()); About                     if(childnum-queue.size () = = 0){ \$ Result.add (node.val); -Childnum = 0; -                     } -                 }  A         } +  the         returnresult; -          \$}`
`1 /**2 * Method Two3 * The solution of others ' home, ingenious, fast and good understanding4      * @paramRoot5 * Right-to-left depth takes precedence over, saving the first node of each layer. Use the size of the current result table as the identity, so that each node that is saved is the first node that appears at that layer6      * @return7      */8      Public   StaticList<integer>rightSideView2 (TreeNode root) {9list<integer> res =NewArraylist<integer>();Ten         if(Root = =NULL){ One             returnRes; A         } -DFS (root, res, 0); -         returnRes; the     } -      -      Public Static voidDFS (TreeNode root, list<integer> res,intLevel ) { -         if(Root = =NULL){ +             return; -         } +         if(res.size () = =Level ) { A Res.add (root.val); at         } -         if(Root.right! =NULL){ -DFS (Root.right, res, level + 1); -         } -         if(Root.left! =NULL){ -DFS (Root.left, res, level + 1); in         } -}`

Leetcode 199:binary Tree Right Side View

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