java fork join

voidDisplayrootlast (node node) {if(Node.leftchild! =NULL) Displayrootlast (node.leftchild); if(Node.rightchild! =NULL) Displayrootlast (node.rightchild); System.out.println ("Node Value:" +node.data); } Public Static voidMain (string[] a) {Linkofbtree tree=NewLinkofbtree (10); Tree.add (8); Tree.add (7); Tree.add (3); Tree.add (11); Tree.add (3); Tree.add (20); Tree.add (26); Tree.add (11); System.out.println ("------Root traversal------"); Tree.displayrootfirst (Tree.root); System.

Binary search tree: is a binary tree in which each node contains a key and a value associated with it, and each node has a key greater than the key of any node in its left subtree, which is less than the key of any node in its right sub-tree.The principle of each method of binary search tree is explained in detail in the code, the following is the code:/** * * @author seabear * Binary search tree * @param Copyright NOTICE: This article for Bo Master original article, without Bo Master permissio

Analysis:Code: PackageTree; Public class verifybst { //First element subscript low, last element subscript high Public Static Boolean Verifysequencebst(int[] sequence,intLowintHigh) {if(Sequence = =NULL|| High-Low 0){return false; }introot = Sequence[high];intIThe node of the left sub-node of the binary search tree is smaller than the root junction . for(i=low;i//If the value is greater than the root node, stop, indicating that the subscript starts with the right subtree .

Topic:Given N, how many structurally unique BST 's (binary search trees) that store values 1 ... n?For example,Given N = 3, there is a total of 5 unique BST ' s. 1 3 3 2 1 \// /\ 3 2 1 1 3 2 / / \ 2 1 2 3Solving:With recursive thinking, when there are only 0 or 1 nodes, there is only one; n nodes have f (n):The left can have n-1 nodes, the right 0 nodes, according to symmetry

Topic:Given N, how many structurally unique BST 's (binary search trees) that store values 1 ... n?For example,Given N = 3, there is a total of 5 unique BST ' s. 1 3 3 2 1 \// /\ 3 2 1 1 3 2 / / \ 2 1 2 3Solving:With recursive thinking, when there are only 0 or 1 nodes. There is only one. N nodes have an F (n) Type:The left side can have n-1 nodes, the right 0 nodes, accordin

-recursive post-traversal") {//need to focus on understanding@Override Public voidExecute (Node node) {FinalstackNewStack(); Node tmp=NULL; while(Node! =NULL) { while(Node.getleft ()! =NULL) {Stack.push (node); Node=Node.getleft (); } while(Node! =NULL (node.getright () = =NULL|| Node.getright () = = tmp)) {//The current node has no right subtree or right subtree output.Visit (node); //record last output nodeTMP =node; if(Stack.empty ())return; Node=Stack.pop ();

In Path SUm 1 (http://www.cnblogs.com/hitkb/p/4242822.html)We save the path in the form of a stack, and each time we find a matching leaf node, the elements in the stack are output. Note the case where multiple paths exist. PublicListintsum) {ListNewArraylist(); StackNewStack(); intTotal = 0; if(Root = =NULL) returnlist; TreeNode Qnode=Root; while(Root! =NULL) { while(Root.left! =NULL) {Stack.push (root); Total+=Root.val; Root=Root.left; }//left node compression stack

Package H1;import Java.lang.thread.state;import Java.util.stack;public class Tree {public static void main (string[] args {Tree tree = new Tree (); Tree.insert (8); Tree.insert (5); tree.insert; Tree.insert (4); Tree.insert (6); Tree.insert (9 ); Tree.insert (n); Tree.postorder (Tree.root); System.out.println (); tree. Posttraverse (tree.root);} Private Node Root;public boolean find (int key) {Boolean flag = true; Node current = Root;while (Current.getidata ()! = key) {if (Key

Title: Please complete a function, enter a binary tree, the function output its mirror The structure of the binary tree is defined as: Package utils; public class Binarytreenode {public int value; Public Binarytreenode Leftnode; Public Binarytreenode Rightnode; Public Binarytreenode () { } public binarytreenode (int value) { this.value = value; This.leftnode = null; This.rightnode = null; } } The image of the tree is a relatively new concept, we may not be able to find out

access to public void print () {if (root! = null) {Root.printnode ();}} The branch node class nodes {private node leftnode;private node Rightnode;private int data;public node (int data) {this.data = data;} Add node public void AddNode (int data) {if (This.data > data) {//Add on left if (This.leftnode = = null) {This.leftnode = new No De (data);} else {this.leftNode.addNode (data);}} else if (this.data Test class of Binary treePackage com.lym.binarytree;/** * Binary Tree Test class * * @author

) {current = Current.leftnode;if (current = = null) {Parent.leftnode = Newnode;return;} } else {current = current.rightnode;if (current = = null) {Parent.rightnode = Newnode;return;}}}} Find public Bitreenode find (int data) {if (root==null) {throw new RuntimeException ("Binary tree is Empty");} Bitreenode Current=root;while (current.data!=data) {if (current.data>data) {Current=current.leftnode;} Else{current=curRent.rightnode;} if (current==null) {return null;}} return current;}}

On the code:Package Com.itany.erchachazhaoshu;public class BinarysearchtreePackage Com.itany.erchachazhaoshu;public class test{public static void Main (string[] args) { Binarysearchtree bt=new Binarysearchtree (); Bt.insert (3); Bt.insert (+); Bt.insert (1); Try { System.out.println ("Max:" +bt.findmax ()); System.out.println ("Max:" +bt.findmin ()); System.out.println (Bt.contains (3)); Bt.remove (+

Programs from Program CreekAgoPreorder binary tree traversal is a classic interview problem about trees. The key to solve this problem are to understand the following: What is preorder? (parent node is processed before it children) Use Stack from the Java Core Library It isn't obvious what preorder for some strange cases. However, if you draw a stacks and manually execute the program, what each element was pushed and popped is obviou

First, define a node class:Package Test;public class Node {private int data;private node Left;private node right;public node (int data) {This.data = d ATA;} public int GetData () {return data;} public void SetData (int data) {this.data = data;} Public Node GetLeft () {return to left;} public void Setleft (Node-left) {this.left = left;} Public Node GetRight () {return to right;} public void Setright (Node right) {this.right = right;}}Two, define an algorithm implementation class:Package Test;publ

"105-construct Binary tree from preorder and inorder traversal (constructed two-forked trees by sequence and sequence traversal)" " leetcode-interview algorithm classic-java Implementation" "All topic Directory Index" Original title Given Preorder and inorder traversal of a, construct the binary tree. 　　Note:You could assume that duplicates does not exist into the tree.The main effect of the topic A binary tree is constructed by giving a sequence o

then executes the logic of the current thread. */ Public classJointest { Public Static voidMain (string[] args) {Try{nojoin Nojoin=NewNojoin (); Nojoin.start (); Nojoin.join ();//JoinSystem.out.println ("Execute to main ....")); } Catch(interruptedexception e) {e.printstacktrace (); } }}To view the join source code: /*** Waits at the most {@codeMillis} milliseconds for this thread to * die. A timeout of {@code0} means to wait forever.

JAVA beginners [9]-mybatis multi-table join query, java-mybatis Summary This section implements a simple demo of Multi-table join queries. The scenario is to query the product category information based on the id and display the product list under the category.I. Mysql Test Data Create a table Category and Product, and

join method is done by calling the thread's wait method to achieve the purpose of synchronization. For example, a thread called the Join method of the B thread, which is equivalent to a thread called the wait method of the B thread, after the call of the wait method of the B thread, a thread will go into a blocking state, the following source:Public final synchronized void

is the implementation of the Join method?In fact, the join method is done by calling the thread's wait method to achieve the purpose of synchronization. For example, a thread called the Join method of the B thread, which is equivalent to a thread called the wait method of the B thread, after the call of the wait method of the B thread, a thread will go into a bl

); }}Else { while(IsAlive ()) { LongDelay = Millis-Now ; if(Delay ) { Break; } wait (delay);now = System.currenttimemillis ()-Base; } }}Lines 12th and 20th should already be clear.And sleep doesn't release the lock.Sleep (long) method does not put locksThe above example shows that join is locked and sleep is not locked.The following example proves another problem with