Java平衡二叉树

最新推荐文章于 2024-09-16 23:32:03 发布

eVA5220

最新推荐文章于 2024-09-16 23:32:03 发布

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CC 4.0 BY-SA版权

文章标签： Java IDEA

本文链接：https://blog.youkuaiyun.com/eVA5220/article/details/83679036

本文介绍了一个二叉树类的实现，包括节点插入、树的大小获取、中序遍历及显示等功能，并实现了将二叉树转换为链表再重新构造为平衡二叉树的过程。

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package com.neu.util;
public class BTNode {
    int data;
    BTNode left;
    BTNode right;
    
    BTNode(){}
    BTNode(int val){
        data = val;
    }
    BTNode(int val,BTNode leftChild,BTNode rightChild){
        data = val;
        left = leftChild;
        right = rightChild;
    }
    public String toString()
    {
        Integer data1= new Integer(data);
        return data1.toString();
    }

}


package com.neu.util;

	public class BTree {
	    
	    public BTNode root;//树的根结点
	    private int count;//树的结点个数
	        
	    /** Creates a new instance of BTree */
	    public BTree() {
	        root = null;
	        count = 0;
	    }
	    public   BTNode insert(BTNode ref,int val)//插入建立二叉树
	    {
	        if(ref == null){
	            count++;
	            return new BTNode(val);
	        }
	        else{
	            if(ref.data>val)
	                ref.left = insert(ref.left,val);
	            else
	                ref.right = insert(ref.right,val);
	            return ref;
	        }
	    }
	    public void insert(int val)
	    {
	        root = insert(root,val);
	    }
	    
	    public int size(){
	        return count;
	    }
	    //将二叉树转为按中序排列的一条链
	    BTNode createChain(BTNode root){
	        if(root!= null){
	            root.right = createChain(root.right); //对右子树递归  
	            if(root.left != null){
	                BTNode oldRoot = root;
	                BTNode ref = root = createChain(root.left); //对左子树递归,根指向最左结点
	                oldRoot.left = null;
	                while(ref.right != null)
	                    ref = ref.right;
	                ref.right = oldRoot;
	            }
	        }
	        return root;
	    } 
	    //将链转为平衡二叉树
	    BTNode createTree(BTNode root,int n){
	        if(n> 2){
	            int m = (n-1)/2;
	            BTNode oldRoot = root;
	            BTNode ref = root;
	            for(int i = 1;i<m;i++)
	                ref = ref.right;
	            root = ref.right;   //指向第[n/2]个结点,即将根指向链的中间结点
	            ref.right = null;  //根的左右结点都置空
	            root.left = createTree(oldRoot,m);
	            root.right = createTree(root.right,n-m-1);
	        }
	        return root;
	    }
	    
	    public String getData() {
	       if(this.root == null)  return "";
	       Integer n = new Integer(root.data);
	       return n.toString();
	    }  
	
	
	public static void inorder(BTNode r){       //中序排列
		if(r != null){
		   inorder(r.left);  
		   System.out.print(r.toString() + "  ");
		   inorder(r.right);
		} 
	}

	public void display(){
		if(root!=null){
			System.out.print("排序：");
			inorder(root);
		}else{
			System.out.println("二叉树为空！");
		}
	}  
	
	public static void main(String[] args){
		BTNode node = new BTNode(10);
		BTree bt =new BTree();   
		int arr[]={5,12,20,2,14,1};
		for(int i=0;i<arr.length;i++)    {    
			bt.insert(arr[i]);    
		 }
		System.out.println();
		bt.display();

	}
}