將計數器存儲在Java中的二叉樹遞歸函數中

Question

我在學校做一個練習，我必須在java中創建一個方法（最多O（n）時間），該方法存儲給定值在二叉樹中出現的順序，無論是后順序，按順序還是按順序預遍歷。 到目前為止，我有有效的代碼，但是計數器不能正確遞增。 有人可以讓我知道我在做什么錯嗎？ 我的代碼如下：

public class Node {

int value;                // data used as key value
Node left;                // this node's left child
Node right;               // this node's right child
int preorderNumber; // node number in preorder
int inorderNumber; // node number in inorder
int postorderNumber; // node number in postorder


public Node(int item) { //instantiates node.
    value = item;
    left = right = null;
    preorderNumber = inorderNumber = postorderNumber = 0;
}

}

/* Given a binary tree, print its nodes according to the postorder traversal. */
private void printPostorder(Node node) {
    int counter = 0;

    if (node == null) { //the tree is empty. 
        return;
    }
    printPostorder(node.left); // first recur on left subtree 
    printPostorder(node.right); // then recur on right subtree
    System.out.print(node.value + " "); // now deal with the node
    counter++;
    node.postorderNumber = counter;
}

/* Given a binary tree, print its nodes in inorder*/
private void printInorder(Node node) {
    int counter = 0;

    if (node == null) {
        return;
    }
    printInorder(node.left);   // first recur on left child
    System.out.print(node.value + " "); // then print the data of node
    counter++;
    node.inorderNumber = counter;
    printInorder(node.right); // now recur on right child 
}

/* Given a binary tree, print its nodes in preorder*/
private void printPreorder(Node node) {
    int counter = 0;

    if (node == null) { //binary tree is empty. 
        return;
    }
    System.out.print(node.value + " "); // first print data of node.
    counter++;
    node.preorderNumber = counter;
    printPreorder(node.left); //  recur on left subtree 
    printPreorder(node.right); // now recur on right subtree 
}

public int find(int valueToFind) { 
    Node current = root, prior = null; 
    while (current != null) {
        int compare;
        compare = (valueToFind - current.value);
        if (compare < 0) {
            prior = current;
            current = current.left;
        } else if (compare > 0) {
            current = current.right;
        } else {
            System.out.println("The node value searched for is : " + valueToFind);
            System.out.println("The in order value is: " + current.inorderNumber);
            System.out.println("The post order value is: " + current.postorderNumber);
            System.out.println("The pre order value is: " + current.preorderNumber);
            return current.value;
        }
    }
    return prior == null ? null : prior.value;
}

}

每當我嘗試運行此命令時，都會得到類似以下輸出的信息：

節點數為：2二叉樹的順序遍歷為3 4二叉樹的順序遍歷為3 4二叉樹的后順序遍歷為4 3搜索的節點值為：4順序值是：1后順序值是：1預定值是：1

我不確定如何將計數器的值傳遞給節點，以便它保留正確的值（在這種情況下，為2、2和1）。 謝謝！

Answer 1

您需要從prv值獲取計數器。 否則，它將一直被重置。 像下面那樣做

private void printPostorder(Node node) {

    if (node == null) { //the tree is empty. 
        return;
    }

    int counter = node.postorderNumber;

    printPostorder(node.left); // first recur on left subtree 
    printPostorder(node.right); // then recur on right subtree
    System.out.print(node.value + " "); // now deal with the node
    counter++;
    node.postorderNumber = counter;
}

預購方法也應該像這樣改變，否則計數器不會增加