二叉排序树的实现

今天实现的是一个二叉排序树的实现，我只是实现了add和delete的方法，本实现是保证在单核的情况下进行设计的，不考虑多线程的并发状态。所以说，还请见谅，有什么疑问或者能更好的实现方法的，可以与我联系，欢迎大家一起进行探讨。

package com.util.support;

public class Tree {

	TreeNode rootNode;

	public void addTree(TreeNode node) {
		if (node == null)
			throw new IllegalArgumentException();
		if (rootNode == null)
			rootNode = node;
		else {
			TreeNode parentNode;
			TreeNode current = rootNode;
			for (;;) {
				parentNode = current;
				if (current.data > node.data) {
					current = current.rightNode;
					if (current == null) {
						parentNode.rightNode = node;
						return;
					}
				} else if (current.data < node.data) {
					current = current.leftNode;
					if (current == null) {
						parentNode.leftNode = node;
						return;
					}
				} else if (current.data == node.data) {
					return;
				}
			}
		}

	}

	public void add(int data)
	{
		if (rootNode == null)
			rootNode = new TreeNode(data);
		else {
			TreeNode parentNode;
			TreeNode current = rootNode;
			for (;;) {
				parentNode = current;
				if (current.data < data) {
					current = current.rightNode;
					if (current == null) {
						parentNode.rightNode = new TreeNode(data);
						return;
					}
				} else if (current.data > data) {
					current = current.leftNode;
					if (current == null) {
						parentNode.leftNode = new TreeNode(data);
						return;
					}
				} else if (current.data == data) {
					return;
				}
			}
		}
	}

	public void delete(int data) {
		NodeInfo info = queryNode(data);

		// 开始对该NodeInfo进行分析
		
		//没有找到
		if(info.nofound)
			return;
		else
		{
			//如果是根节点
			if(info.parentNode == null)
			{
				//根节点不含有左孩子和右孩子
				if(rootNode.leftNode == rootNode.rightNode)
				{
					rootNode = null;
				}else if(rootNode.leftNode == null)//左孩子为空
				{
					TreeNode temp  = rootNode;
					
					rootNode  = rootNode.rightNode;
					
					temp.rightNode  = null;
				}else if(rootNode.rightNode == null)
				{
					TreeNode temp  = rootNode;
					
					rootNode = rootNode.leftNode;
					
					temp.leftNode = null;
				}else//这一种最为复杂，我们找到左子树中最大的节点，将其进行代替。
				{
					deleteNodeSpecial(rootNode);
				}
			}else{
				
				TreeNode node = info.left ? info.parentNode.leftNode:info.parentNode.rightNode;
				
				if(node.leftNode == node.rightNode)
				{
					if(info.left) info.parentNode.leftNode = null;
					else info.parentNode.rightNode = null;
				}else if(node.leftNode == null)
				{
					if(info.left) info.parentNode.leftNode = node.rightNode;
					else info.parentNode.rightNode = node.rightNode;
					
					node.rightNode = null;
					
				}else if(node.rightNode == null)
				{
					if(info.left) info.parentNode.leftNode = node.leftNode;
					else info.parentNode.rightNode  = node.leftNode;
					
					node.leftNode = null;
				}else {
					deleteNodeSpecial(node,info);
				}
			}
		}
	}

	private void deleteNodeSpecial(TreeNode node,NodeInfo info) {
		TreeNode parentNode = node;
		TreeNode current    = parentNode.leftNode;
		while(current.rightNode != null)
		{
			parentNode  = current;
			
			current  = current.rightNode;
		}
		current.rightNode = node.rightNode;
		
		if(info.left)	 info.parentNode.leftNode = node.leftNode;
		else info.parentNode.rightNode = node.leftNode;
		
		node.leftNode = null;
		node.rightNode = null;
	}

	private void deleteNodeSpecial(TreeNode node) {
		TreeNode parentNode = node;
		TreeNode current    = parentNode.leftNode;
		while(current.rightNode != null)
		{
			parentNode  = current;
			
			current  = current.rightNode;
		}
		
		//找到最大的节点 current
		parentNode.rightNode = null;
		//该节点还有左孩子
		if(current.leftNode != null)
		{
			parentNode.rightNode = current.leftNode;
			current.leftNode = null;
		}
		//进行代替
		node.data = current.data;
	}
	
	private NodeInfo queryNode(int data) {
		
		NodeInfo info = new NodeInfo();
		
		/**
		 * 是不是头节点
		 */
		if (rootNode.data == data) {
			info.left = false;
			info.parentNode = null;
			return info;
		}

		TreeNode parentNode = rootNode;

		for (;;) {
			if (parentNode.data < data) {
				
				//没有找到
				if(parentNode.rightNode  == null){
					info.nofound = true;
					return info ;
				}else if(parentNode.rightNode.data  == data){
					info.left  = false;
					info.nofound  = false;
					info.parentNode = parentNode;
					return info;
				}else{
					parentNode  = parentNode.rightNode;
				}
			}else{
				if(parentNode.leftNode == null)
				{
					info.nofound  = true;
					return info;
				}else if(parentNode.leftNode.data == data)
				{
					info.left = true;
					info.nofound = false;
					info.parentNode = parentNode;
					
					return info;
				}else{
					parentNode = parentNode.leftNode;
				}
			}
		}
	}

	/**
	 * 记录一个Node的Info
	 * 
	 * @author Administrator
	 * 
	 */
	private final class NodeInfo {
		TreeNode parentNode;

		boolean left = false;
		
		boolean nofound  = false;	
	}

	public TreeNode getRootNode() {
		return rootNode;
	}
}

下面是测试的代码，测试的思想是，我输入无序的数组，依次进行输入到树中，最后如果tree在add完之后，我们使用的是中序的遍历方法，如果中序的遍历方法是有序的话，证明我们的add的方法是正确的。同理，如果我们随意删除一个节点的值，最后进行中序遍历，如果仍然是有序的证明我们仍然正确。

public static void main(String[] args) {

		Tree tree  = new Tree();
		
		int[] arrays = new int[] { 128, 137, 145, 55, 532, 99, 12, 3, 1000, 255 };
		
		for(int data : arrays)
		{
			tree.add(data);
		}
		
		//System.out.println(tree.getRootNode().getLeftNode().getData());
		
		printBinaryTree(tree.getRootNode());
		
		System.out.println();
		
		tree.delete(532);
		
		printBinaryTree(tree.getRootNode());
	}
	
	/**
	 * 中序遍历二叉树 那么这个二叉树应该是有序的序列
	 * @param tree
	 */
	private static void printBinaryTree(TreeNode root) {
		
		if(root != null)
		{
			printBinaryTree(root.getLeftNode());
			System.out.print(root.getData() + " ");
			printBinaryTree(root.getRightNode());
		}
		
	}