Java Collection Framework Binary Tree
Example 1: Binary tree concept
A binary tree consists of nodes
Binary Tree Features:
Each node can have left and right child nodes
Each node has a value
package collection; public class Node { // Left Child Node public Node leftNode; // Right Child Node public Node rightNode; // value public Object value; }
Example 2: Binary tree sorting - inserting data
Suppose you sort the following 10 random numbers by a binary tree
67,7,30,73,10,0,78,81,10,74
The first step in sorting is to insert data into the binary tree
The basic logic of insertion is that small, the same, are placed on the left and large on the right
- 67 at the root node
- 7 smaller than 67, placed on the left node of 67
- 30 is smaller than 67, find the left node of 67 7, 30 is larger than 7, put it on the right node of 7
- 73 is larger than 67, placed on the right node of 67
- 10 is smaller than 67, find 67 left node 7, 10 is larger than 7, find 7 right node 30, 10 is smaller than 30, put in 30 left node.
...
... -
10 is smaller than 67, find 67 left node 7, 10 is larger than 7, find 7 right node 30, 10 is smaller than 30, find 30 left node 10, 10 is the same size as 10, put on the left
package collection; public class Node { // Left Child Node public Node leftNode; // Right Child Node public Node rightNode; // value public Object value; // Insert Data public void add(Object v) { // If the current node has no value, place the data on the current node if (null == value) value = v; // If the current node has a value, make a judgment about the size of the new value in relation to the current value else { // New value, less than or equal to the current value if ((Integer) v -((Integer)value) <= 0) { if (null == leftNode) leftNode = new Node(); leftNode.add(v); } // New value, greater than current value else { if (null == rightNode) rightNode = new Node(); rightNode.add(v); } } } public static void main(String[] args) { int randoms[] = new int[] { 67, 7, 30, 73, 10, 0, 78, 81, 10, 74 }; Node roots = new Node(); for (int number : randoms) { roots.add(number); } } }
Example 3: Binary tree sorting-traversal
With the insertion behavior of the previous step, the data is actually sorted.The next thing to do is to iterate through these sorted data into lists or arrays that we commonly use
Traversal of a binary tree is in left, middle, and right order
Left order means that the middle number is traversed and placed on the left
The middle order is: the middle number is traversed and placed in the middle
The right order is: the middle number is traversed and placed on the right
As you can see from the diagram, we want the results to be small to large, so we should use a middle-order traversal
package collection; import java.util.ArrayList; import java.util.List; public class Node { // Left Child Node public Node leftNode; // Right Child Node public Node rightNode; // value public Object value; // insert data public void add(Object v) { // If the current node has no value, place the data on the current node if (null == value) value = v; // If the current node has a value, make a judgment about the size of the new value in relation to the current value else { // New value, less than or equal to the current value if ((Integer) v -((Integer)value) <= 0) { if (null == leftNode) leftNode = new Node(); leftNode.add(v); } // New value, greater than current value else { if (null == rightNode) rightNode = new Node(); rightNode.add(v); } } } // Traverse all nodes in middle order public List<Object> values() { List<Object> values = new ArrayList<>(); // Traversal result of left node if (null != leftNode) values.addAll(leftNode.values()); // Current Node values.add(value); // Traversal result of right node if (null != rightNode) values.addAll(rightNode.values()); return values; } public static void main(String[] args) { int randoms[] = new int[] { 67, 7, 30, 73, 10, 0, 78, 81, 10, 74 }; Node roots = new Node(); for (int number : randoms) { roots.add(number); } System.out.println(roots.values()); } }
Practice: Hero Binary Tree
Based on the above learning and understanding, design a Hero binary tree, HeroNode.
You can insert different Hero objects into this Hero Binary Tree and sort them in reverse order according to Hero's blood volume.
Ten Hero objects were randomly generated, each with a different blood volume. After inserting this HeroNode, the sorting results were printed.
package collection; import java.util.ArrayList; import java.util.List; import charactor.Hero; public class HeroNode { public HeroNode leftHero; public HeroNode rightHero; // Declare as Hero type public Hero value; public void add(Hero v) { if (null == value) value = v; else { // If the new hero has more blood than this node, put it on the left if (v.hp > value.hp) { if (null == leftHero) leftHero = new HeroNode(); leftHero.add(v); } else { if (null == rightHero) rightHero = new HeroNode(); rightHero.add(v); } } } public List<Object> values() { List<Object> values = new ArrayList<>(); if (null != leftHero) values.addAll(leftHero.values()); values.add(value); if (null != rightHero) values.addAll(rightHero.values()); return values; } public static void main(String[] args) { List<Hero> hs = new ArrayList<>(); for (int i = 0; i < 10; i++) { Hero h = new Hero(); h.name = "hero " + i; h.hp = (float) (Math.random() * 900 + 100); // 100-1000 Random Blood Volume hs.add(h); } System.out.println("Initialize 10 Hero"); System.out.println(hs); HeroNode heroTree = new HeroNode(); for (Hero hero : hs) { heroTree.add(hero); } System.out.println("Reversed by blood volume Hero"); List<Object> treeSortedHeros = heroTree.values(); System.out.println(treeSortedHeros); } }