具有遍歷的Java中的二進制搜索樹實現

Question

我已經實現了帶有插入和遍歷方法的二叉搜索樹，但是沒有為PreOrder和Postorder獲得正確的輸出，而是以正確的順序獲得inOrder。 有人可以告訴我哪里錯了。

我在紙上嘗試了相同的示例，但PreOrder和PostOrder不同。

這是我的代碼

節點類別

package com.BSTTest;

public class Node implements Comparable<Node> {
    private int data;
    private Node leftChild;
    private Node rightChild;

    public Node(int data) {
        this(data, null, null);
    }

    public Node(int data, Node leftChild, Node rightChild) {
        this.data = data;
        this.leftChild = leftChild;
        this.rightChild = rightChild;

    }

    public int getData() {
        return data;
    }

    public void setData(int data) {
        this.data = data;
    }

    public Node getLeftChild() {
        return leftChild;
    }

    public void setLeftChild(Node leftChild) {
        this.leftChild = leftChild;
    }

    public Node getRightChild() {
        return rightChild;
    }

    public void setRightChild(Node rightChild) {
        this.rightChild = rightChild;
    }

    public int compareTo(Node o) {
        return Integer.compare(this.data, o.getData());
    }
}

樹類

package com.BSTTest;

import com.BSTTest.Node;

public class Tree {
    private Node root = null;

    public Node getRoot() {
        return root;
    }
     //Inserting data**strong text**
    public void insertData(int data) {
        Node node = new Node(data, null, null);
        if (root == null) {
            root = node;
        } else {
            insert(node, root);
        }
    }
    //Helper method for insert
    private void insert(Node node, Node currNode) {
        if (node.compareTo(currNode) < 0) {
            if (currNode.getLeftChild() == null) {
                currNode.setLeftChild(node);
            } else {
                insert(node, currNode.getLeftChild());
            }
        } else {

            if (currNode.getRightChild() == null) {
                currNode.setRightChild(node);
            } else {
                insert(node, currNode.getRightChild());
            }
        }
    }

    public void printInorder() {
        printInOrderRec(root);
        System.out.println("");
    }


      //Helper method to recursively print the contents in an inorder way

    private void printInOrderRec(Node currRoot) {
        if (currRoot == null) {
            return;
        }
        printInOrderRec(currRoot.getLeftChild());
        System.out.print(currRoot.getData() + ", ");
        printInOrderRec(currRoot.getRightChild());
    }

    public void printPreorder() {
        printPreOrderRec(root);
        System.out.println("");
    }


     // Helper method for PreOrder Traversal recursively 

    private void printPreOrderRec(Node currRoot) {
        if (currRoot == null) {
            return;
        }
        System.out.print(currRoot.getData() + ", ");
        printPreOrderRec(currRoot.getLeftChild());
        printPreOrderRec(currRoot.getRightChild());
    }

    public void printPostorder() {
        printPostOrderRec(root);
        System.out.println("");
    }

    /**
     * Helper method for PostOrder method to recursively print the content
     */
    private void printPostOrderRec(Node currRoot) {
        if (currRoot == null) {
            return;
        }
        printPostOrderRec(currRoot.getLeftChild());
        printPostOrderRec(currRoot.getRightChild());
        System.out.print(currRoot.getData() + ", ");
    }
    //Main Mthod
    public static void main(String[] args) {
        Tree obj = new Tree();
        //Inserting data
        obj.insertData(3);
        obj.insertData(5);
        obj.insertData(6);
        obj.insertData(2);
        obj.insertData(4);
        obj.insertData(1);
        obj.insertData(0);

        //printing content in Inorder way
        System.out.println("Inorder traversal");
        obj.printInorder();

        //printing content in Inorder way
        System.out.println("Preorder Traversal");
        obj.printPreorder();

        //printing content in Inorder way
        System.out.println("Postorder Traversal");
        obj.printPostorder();
    }
}

Answer 1

朋友，您的代碼絕對正確，因為您提到的輸出絕對正確。

我認為您沒有正確理解Binary Search Tree的概念。

您是正確的，3是根節點，但您說1是它的左子節點是錯誤的。

在3之后出現且小於3的第一個值是2，因此2是3的子級，而不是1

如果仍然有困惑，請參閱Cormenn書。