java删除二叉树节点的实现
想求懂java的大哥指导为什我写的del方法无法删除节点BinaryTree类创建二叉树:
public class BinaryTree {
private Hero root;
Scanner myScanner =new Scanner(System.in);
public Hero createBinary() {
Hero hero = null;
System.out.println("你是否需要创造新节点y/n");
String key = myScanner.next();
if ("y".equals(key)) {
System.out.print("请输入 id = ");
int id = myScanner.nextInt();
System.out.print("请输入 name = ");
String name = myScanner.next();
hero = new Hero(id, name);
} else if ("n".equals(key)) {
// hero = null;
return null;
}
hero.setLeftChild(createBinary());
hero.setRightChild(createBinary());
return hero;
}
// Hero p = root;
// if (p == null) {
// root = hero;
// return;
// } else {
// if (p.getLeftChild() == null) {
// p.setLeftChild(hero);
// p = p.getLeftChild();
// } else if (p.getRightChild() == null) {
// p.setRightChild(root);
// p = p.getRightChild();
// }
// }
public Hero getRoot() {
return root;
}
public void setRoot(Hero root) {
this.root = root;
}
}
Hero是节点类
public class Hero {
private Hero leftChild;
private Hero rightChild;
private int id;
private String name;
private int tag = 0;//0为未遍历，1为已遍历
public int getTag() {
return tag;
}
public void setTag(int tag) {
this.tag = tag;
}
public Hero(int id, String name) {
this.id = id;
this.name = name;
}
public Hero getLeftChild() {
return leftChild;
}
public void setLeftChild(Hero leftChild) {
this.leftChild = leftChild;
}
public Hero getRightChild() {
return rightChild;
}
public void setRightChild(Hero rightChild) {
this.rightChild = rightChild;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "Hero{" +
"id=" + id +
", name='" + name + '\'' +
'}';
}
}
DeleteTree是实现删除节点的类
public class DeleteTree {
private static Scanner myScanner = new Scanner(System.in);
public static void del(Hero tree, int id) {
Stack<Hero> stack = new Stack<>();
Hero p = tree;
if (p == null) {
System.out.println("该二叉树为空...");
return;
} else {
while (p != null) {
if (p.getId() == id) {
p = null;
}
else {
if (p.getRightChild() != null) {
stack.push(p.getRightChild());
}
if (p.getLeftChild() != null) {
p = p.getLeftChild();
} else if (!stack.empty()) {
p = stack.pop();
} else {
break;
}
}
}
}
}
}
TraMain是实现main方法的类
public class TraMain {
static Scanner myScanner = new Scanner(System.in);
public static void main(String[] args) {
TraMain traMain = new TraMain();
BinaryTree binaryTree = new BinaryTree();
Hero BinaryTree = binaryTree.createBinary();//这样子一颗二叉树就搞好了
System.out.println("===============递归前序遍历===============");
traMain.preTravers(BinaryTree);//前序遍历
DeleteTree.del(BinaryTree, 2);
System.out.println("===============递归前序遍历===============");
traMain.preTravers(BinaryTree);//前序遍历
/**
System.out.println("===============递归中序遍历===============");
traMain.inorderTra(BinaryTree);
System.out.println("===============递归后序遍历===============");
traMain.postTra(BinaryTree);
System.out.println("===============非递归前序遍历===============");
traMain.nonRePreTra(BinaryTree);
System.out.println("===== 中序非递归遍历(order) / 后序非递归遍历(post) =====");
System.out.print("请输入 order 或 post: ");
String key = myScanner.next();
if ("order".equals(key)) {
System.out.println("===============非递归中序遍历===============");
traMain.nonReorderTra(BinaryTree);
} else {
System.out.println("===============非递归后序遍历===============");
traMain.nonRepostTra(BinaryTree);
}
**/
}
//前序遍历二叉树的方法
public void preTravers(Hero tree) {
Hero p = tree;
if (p == null) {
return;
}
System.out.println(p);
preTravers(p.getLeftChild());
preTravers(p.getRightChild());
}
public void inorderTra(Hero tree) {
Hero p = tree;
if (p == null) {
return;
}
inorderTra(p.getLeftChild());
System.out.println(p);
inorderTra(p.getRightChild());
}
public void postTra(Hero tree) {
Hero p = tree;
if (p == null) {
return;
}
postTra(p.getLeftChild());
postTra(p.getRightChild());
System.out.println(p);
}
public void nonRePreTra(Hero tree) {
Stack<Hero> stack = new Stack<Hero>();
Hero p = tree;
if (p == null) {
return;
}
while (p != null) {
if (p.getRightChild() != null) {
stack.push(p.getRightChild());
}
System.out.println(p);
if (p.getLeftChild() != null) {
p = p.getLeftChild();
} else if (stack.size() > 0) {
p = stack.pop();
} else {
p = null;
}
}
}
public void nonReorderTra(Hero tree) {
Stack<Hero> stack = new Stack<Hero>();
Hero p = tree;
if (p == null) {
return;
}
while (p != null) {
while (p.getLeftChild() != null && p.getLeftChild().getTag() == 0) {
stack.push(p);
p = p.getLeftChild();
}
System.out.println(p);
p.setTag(1);
if (p.getRightChild() != null) {
p = p.getRightChild();
} else if (stack.size() != 0){
p = stack.pop();
} else {
p = null;
}
}
}
public void nonRepostTra(Hero tree) {
Stack<Hero> stack = new Stack<Hero>();
Hero p = tree;
if (p == null) {
return;
}
while (p != null) {
while (p.getLeftChild() != null && p.getLeftChild().getTag() == 0) {
stack.push(p);
p = p.getLeftChild();
}
if (p.getRightChild()!= null && p.getRightChild().getTag() == 0) {
stack.push(p);
p = p.getRightChild();
} else {
System.out.println(p);
p.setTag(1);
if (stack.size() != 0) {
p = stack.pop();
} else if (stack.size() == 0) {
p = null;
}
}
}
}
}
很高兴能够帮助您。按照您所提供的代码，我看不到有实现删除节点的代码行或方法调用。您可能需要添加以下代码：
public void delNode(Hero root, int id) {
if (root == null) {
return;
}
if(root.getId() == id){
root = null;
}else{
delNode(root.getLeftChild(),id);
delNode(root.getRightChild(),id);
}
}
以上方法将递归遍历二叉树，搜索要删除的节点，并将其设置为NULL。
此外，在以下代码中：
DeleteTree.del(BinaryTree, 2);
应该让您想删除哪个节点作为“2”。这是一个静态调用，因此如果无法直接修复，则可能会出现nullPointerException。
最后，建议您在指向对象之前检查它是否为零。这样做可以避免NullPointerException异常。
我希望这可以解决您的问题！
有用请设置最佳答案
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