想问写java的大师想为什么我删除节点的方法为什么实现不了
BinaryTree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删除节点的方法
public class DeleteTree {
private static Scanner myScanner = new Scanner(System.in);
public static void del(Hero tree, int id) {
Stack 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实现main方法在这实现测试del方法
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 stack = new Stack();
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 stack = new Stack();
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 stack = new Stack();
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;
}
}
}
}
}
你的代码中有一些问题需要修正。首先，在二叉树创建方法 createBinary() 中，当用户输入 'n' 时，应该返回 null 而不是赋值为 null。其次，在节点删除方法 del() 中，你只是设定要删除的节点为空，但并没有实际上删除它，所以需要对删除操作进行改进。
以下是修改建议：
BinaryTree类中修改如下：
public Hero createBinary() {
// your code...
} else if ("n".equals(key)) {
return null;
}
// your code...
}
DeleteTree类中修改如下：
public static void del(Hero tree, int id) {
if (tree == null) {
System.out.println("该二叉树为空...");
return;
}
Hero p = tree;
Hero parent = null;
boolean isLeftChild = true;
while (p != null && p.getId() != id) {
parent = p;
if (id < p.getId()) {
isLeftChild = true;
p = p.getLeftChild();
} else {
isLeftChild = false;
p = p.getRightChild();
}
}
if (p == null) {
System.out.println("未找到待删除节点");
return;
}
// 如果待删除节点是叶子节点
if (p.getLeftChild() == null && p.getRightChild() == null) {
if (p == tree) {
tree = null;
} else if (isLeftChild) {
parent.setLeftChild(null);
} else {
parent.setRightChild(null);
}
}
// 如果待删除节点只有左子节点
else if (p.getRightChild() == null) {
if (p == tree) {
tree = p.getLeftChild();
} else if (isLeftChild) {
parent.setLeftChild(p.getLeftChild());
} else {
parent.setRightChild(p.getLeftChild());
}
}
// 如果待删除节点只有右子节点
else if (p.getLeftChild() == null) {
if (p == tree) {
tree = p.getRightChild();
} else if (isLeftChild) {
parent.setLeftChild(p.getRightChild());
} else {
parent.setRightChild(p.getRightChild());
}
}
// 如果待删除节点有两个子节点
else {
Hero successor = getSuccessor(p);
if (p == tree) {
tree = successor;
} else if (isLeftChild) {
parent.setLeftChild(successor);
} else {
parent.setRightChild(successor);
}
successor.setLeftChild(p.getLeftChild());
}
}
/**
* 获取待删除节点的后继节点
*/
private static Hero getSuccessor(Hero delNode) {
Hero successorParent = delNode;
Hero successor = delNode;
Hero current = delNode.getRightChild();
while (current != null) {
successorParent = successor;
successor = current;
current = current.getLeftChild();
}
if (successor != delNode.getRightChild()) {
successorParent.setLeftChild(successor.getRightChild());
successor.setRightChild(delNode.getRightChild());
}
return successor;
}
这里建议你创建一个新的 Hero 类型的引用来保存删除节点的父节点，这个应该在删除节点时比较容易处理。同时，如果要删除的节点有左右子节点时，需要寻找它的后继节点来替代它，保证删除后仍然是一棵二叉树。
你的代码缺乏条理，tree node tree的操作，调用代码应该分开。
这是我修改之可以删除根节点以为的所有子节点，现在在想怎么处理根节点
public void del(Hero tree, int id) {
Stack<Hero> stack = new Stack<>();
Stack<Hero> parentStack = new Stack<>();
Hero parent = null;
Hero p = tree;
boolean isLeftChild = true;
if (p == null) {
System.out.println("该二叉树为空...");
return;
} else {
while (p != null && p.getId() != id) {
parent = p;
if (p.getLeftChild() == null && p.getRightChild() == null && !parentStack.empty()) {
parent = parentStack.pop();
}
if (p.getLeftChild() != null && p.getLeftChild().getId() == id) {
p = p.getLeftChild();
} /*else if (p.getRightChild() != null && p.getRightChild().getId() == id) {
///p = p.getLeftChild();
} */else {
if (p.getLeftChild() != null) {
parentStack.push(parent);
if (p.getRightChild() != null) {
stack.push(p.getRightChild());
}
p = p.getLeftChild();
} else if (p.getRightChild() != null){
p = p.getRightChild();
} else if (!stack.empty()) {
p = stack.pop();
} else {
break;
}
/*
if (p.getRightChild() != null) {
stack.push(tree.getRightChild());
}
if (p.getLeftChild() != null) {
p = p.getLeftChild();
} else if (!stack.empty()) {
p = stack.pop();
} //else {
// break;
//}
*/
}
}
if (p.getId() == id && parent.getLeftChild() == p) {
parent.setLeftChild(null);
} else if (p.getId() == id && parent.getRightChild() == p) {
parent.setRightChild(null);
} else {
System.out.println(id + " 节点不存在...");
}
}
}