Question

我接到了这个电话

productTree.displayNode(productTree.newsearchTree(search, productTree.Root()));

调用这些功能

void Tree::displayNode(Node* node) {
    if (node != NULL) {
        cout << node->productdata << endl;
    }
}

Node* Tree::newsearchTree(string search, Node* node) {
    if (node != NULL) {
        newsearchTree(search, node->left);
            if (search == node->productdata.getName()) {
                return node;
            }
        newsearchTree(search, node->right);
    }
}

Node* Root() { return root; };

在尝试显示返回的节点时会产生垃圾输出。

然而，我可以通过这种方式遍历我的树并显示它：

这个电话：

productTree.Inorder(productTree.Root());

功能：

void Tree::Inorder(Node* node) {
    if (node != NULL) {
        Inorder(node->left);
        displayNode(node);
        Inorder(node->right);
    }
}

我感到困惑的是它们有何不同？

编辑：感谢您的帮助，我修改了以下代码，现在工作正常：

Node* Tree::newsearchTree(string search, Node* node) {
    if (node != NULL) {
        newsearchTree(search, node->left);
            if (search == node->productdata.getName()) {
                return node;
            }
        newsearchTree(search, node->right);
    }
    return NULL;
}

void Tree::displayNode(Node* node) {
    if (node != NULL) {
        cout << node->productdata << endl;
    }
    else if (node == NULL) {
        cout << "Product not found" << endl;
    }
}

是的，那还是坏了。这终于修好了我相信：

Node* Tree::newsearchTree(string search, Node* node) {
    if (node != NULL) {
        if (search == node->key) {
            return node;
        }
        if (search < node->key) {
            return newsearchTree(search, node->left);
        }
        else {
            return newsearchTree(search, node->right);
        }
    }
    else return NULL;
}

Answer 1

让我在内联添加注释，以明确你的Tree :: newsearchTree做了什么。

Node* Tree::newsearchTree(string search, Node* node) {
  if (node != NULL) {
    newsearchTree(search, node->left);  // burn CPU but leaves no side effect
    if (search == node->productdata.getName()) {
      return node;  // return node only if the root node equals 'search'
    }
    newsearchTree(search, node->right);  // burn CPU without side effect
  }
  // In majority of the cases (root node != 'search'): return nothing.
}

因为通常newsearchTree（）什么都不返回，所以当你调用productTree.displayNode（productTree.newsearchTree（...））时，你基本上都会打印垃圾。

二进制搜索树两个产生垃圾的类似功能

1 个答案: