我们正在编写一个项目，我们使用链接列表在BST中添加和减去数字。但是，我们可以添加数字但无法删除它们。有人可以查看我们的删除功能，看看有什么不对吗？
// 2016 Spring Data Structure BST Sample Code at MIU (For the project #2)
// You may add/modify the classes but should maintain the basic structure of the classes.
import java.util.*;
import java.util.*;
class Node {
int data; // data
Node lc; //left child pointer
Node rc; //right child pointer
Node(int data){ // constructor
this.data=data;
lc=null;
rc=null;
}
}
class BST {
Node root;
BST(){
root=null;
}
void insert(int newData){ // This is not the full version
Node newNode=new Node(newData);
if(root==null) root=newNode; // root was null
else { // root was not null, so find the place to attach the new node
Node ptr=root; // Node pointer
while(true){
if(newNode.data==ptr.data) {
System.out.println("The node already exist.");
break;
}
else if(newNode.data<ptr.data) {
if(ptr.lc!=null) ptr=ptr.lc;
else {
ptr.lc=newNode;
System.out.println(newNode.data+" was successfully inserted.");
break;
}
}
else if(newNode.data>ptr.data) {
if(ptr.rc!=null) ptr=ptr.rc;
else {
ptr.rc=newNode;
System.out.println(newNode.data+" was successfully inserted.");
break;
}
}
}
}
}
Node delete(Node root,int key){ // delete the nodes
if (root == null) return null;
if (root.data > key) {
root.lc = delete(root.lc, key);
}
else if (root.data < key) {
root.rc = delete(root.rc, key);
}
return null;
}
boolean search(Node root,int key){ // if the search is successful return the Node or return null
if(root==null){
return false;
}
if(root.data== key) {
return true;
}
if(root.data!= key){
return false;
}
return true;
}
int number_nodes(Node n){ // the counting left child
if(n == null)
return 0;
if(n.lc ==null && n.rc==null)
return 1;
else
return 1+number_nodes(n.lc);
}
int rnumber_nodes(Node n){ // the counting right child
if(n == null)
return 0;
if(n.lc ==null && n.rc==null)
return 1;
else
return 1+number_nodes(n.rc);
}
void inorder(Node n){ // recursive inorder travelsal
if(n==null) return;
System.out.print("[");
inorder(n.lc);
System.out.print(n.data);
inorder(n.rc);
System.out.print("]");
}
void preorder(Node n){ // recursive preorder travelsal
if(n==null) return;
System.out.print("[");
System.out.print(n.data);
preorder(n.lc);
preorder(n.rc);
System.out.print("]");
}
void postorder(Node n){ // recursive postorder travelsal
if(n==null) return;
System.out.print("[");
postorder(n.lc);
postorder(n.rc);
System.out.print(n.data);
System.out.print("]");
}
}
public class BST_2015134 { // change the name into your IDs
public static void main(String[] args) {
BST bst=new BST();
Scanner sScan=new Scanner(System.in); // menu
Scanner iScan=new Scanner(System.in); // data
while(true){
System.out.print("\n(q)uit,(i)nsert,(d)elete,(s)earch,i(n)order,(p)reorder,p(o)storder,(h)ow many:");
String uChoice=sScan.next();
if(uChoice.equalsIgnoreCase("i")){
System.out.print("Enter a number to insert:");
int uData=iScan.nextInt();
bst.insert(uData);
}
else if(uChoice.equalsIgnoreCase("d")){ // deletion
System.out.print("enter the delete number");
Scanner s=new Scanner(System.in); // to use new scanner
int delete_num=s.nextInt(); //
bst.delete(bst.root,delete_num);
}
else if(uChoice.equalsIgnoreCase("s")){ // search
System.out.print("enter the search number");
Scanner s=new Scanner(System.in);
int search_num=s.nextInt();
if(bst.search(bst.root,search_num)){
while(true){
System.out.println(" your number is found"); // to tell you your # is found or not found
break;
}
}
else{
System.out.println(" your number is not found");
}
}
else if(uChoice.equalsIgnoreCase("n")){ // in order
bst.inorder(bst.root);
}
else if(uChoice.equalsIgnoreCase("p")){ // pre order
bst.preorder(bst.root);
}
else if(uChoice.equalsIgnoreCase("o")){ // post order
bst.postorder(bst.root);
}
else if(uChoice.equalsIgnoreCase("h")){ // how many
int x,y;
x=bst.number_nodes(bst.root);
y=bst.rnumber_nodes(bst.root);
int total=x+y;
System.out.print(total);
}
if(uChoice.equalsIgnoreCase("q")) break; // quit
}
}
}
答案 0 :(得分：1)
回顾一下您使用此功能所做的事情：
如果预期该键包含在左子树中，则从左子树中删除if是否为递归。如果预期密钥将包含在右子树中，请从左子树中删除if with recursion。
到现在为止还挺好。但是，每次都返回null，因此您所做的所有工作实际上都已丢失。
您还需要处理节点数据等于给定键的情况。在这种情况下，您需要以某种方式重组树。
只要您要删除的节点不是树的根目录，下面的代码就可以执行您想要的操作（您应该为此添加特殊情况）
Node delete(Node root,int key){ // delete the nodes
if (root == null) return null;
if (root.data > key) {
root.lc = delete(root.lc, key);
}
else if (root.data < key) {
root.rc = delete(root.rc, key);
}else{
if(root.rc == null)
return root.lc;
if(root.lc == null)
return root.rc;
//if both the left and right child are effective, you need to restructure the tree
int minData = findMinimum(root.rc);
Node newNode = new Node(minData);
newNode.lc = root.lc;
newNode.rc = delete(root.rc, minData);
return newNode;
}
return root;
}
private int findMinimum(Node root){
if (root.lc == null)
return root.data;
else
return findMinimum(root.lc);
}
请注意，从BST中删除节点并非易事。有关详细信息，请查看this链接。
另请注意，这棵树的效率不高。平衡的树会更好。