((TreeNode<K,V>)e).split(this, newTab, j, oldCap)是干嘛

的？

这段代码是Java中HashMap扩容时调用的split方法，作用是将哈希表中某个桶中的链表或红黑树节点分裂成两个链表或红黑树节点，并将它们分别放入新的哈希表中的对应桶中。

具体来说，这段代码中的参数含义如下：

- e：表示当前节点，即需要被分裂的节点；
- newTab：表示新的哈希表，即扩容后的哈希表；
- j：表示当前节点在旧哈希表中的桶位置；
- oldCap：表示旧哈希表的容量。

该方法会在对当前节点进行分裂的同时，顺便将其后继节点也进行分裂，最终返回分裂后的新节点。

相关问题

HashMap中有句源码TreeNode<K,V> x = this, 这么写好像不报错, 那么我这样写为什么报错呢? class Solution<K,V> { static final class TreeNode<K,V>{} public void test(String[] args) { TreeNode<K, V> n = this; } }

在你给出的代码中，`class Solution<K,V>`是泛型类，`,V>`是`Solution类内部的静态内部类。在`test`方法中，你尝试将`this`赋值`TreeNode<K,V>`类型的变量`n`。

然而，这段代码会报错，原因是泛型类型参数`K`和`V`在`test`方法中没有被明确指定。在Java中，泛型类型参数在使用时需要进行具体化。因此，在你的代码中，编译器无法确定`K`和`V`的具体类型，因此无法将`this`赋值给`TreeNode<K,V>`类型的变量。

要修复这个问题，你可以在调用`test`方法时明确指定具体的类型参数，例如：

Solution<String, Integer> solution = new Solution<>();
solution.test(args);

在这个例子中，我们为`K`指定了`String`类型，为`V`指定了`Integer`类型，使得编译器能够正确地进行类型推断和匹配。

总结来说，报错的原因是泛型类型参数未被具体化。为了避免这个错误，需要在使用泛型类或方法时明确指定具体的类型参数。

代码改进，不使用cout<<endl结构#include <iostream> #include <queue> #include <stack> #include <string> using namespace std; struct Student { string name; int number; int score; }; struct TreeNode { Student data; TreeNode* left; TreeNode* right; TreeNode(Student s) : data(s), left(nullptr), right(nullptr) {} }; void postOrder(TreeNode* root) { if (!root) return; postOrder(root->left); postOrder(root->right); cout << root->data.name << " " << root->data.number << " " << root->data.score << endl; } void preOrder(TreeNode* root) { stack<TreeNode*> s; s.push(root); while (!s.empty()) { TreeNode* node = s.top(); s.pop(); if (node) { cout << node->data.name << " " << node->data.number << " " << node->data.score << endl; s.push(node->right); s.push(node->left); } } } void levelOrder(TreeNode* root) { queue<TreeNode*> q; q.push(root); while (!q.empty()) { TreeNode* node = q.front(); q.pop(); if (node) { cout << node->data.name << " " << node->data.number << " " << node->data.score << endl; q.push(node->left); q.push(node->right); } } } int main() { Student students[] = { {"Lei Zhenzi", 101401, 82}, {"Jiang Ziya", 100032, 90}, {"Ne Zha", 101674, 70}, {"Shen Gongbao", 101982, 87}, {"Jiu Weihu", 107431, 75}, {"Tian Zun", 100001, 98}, {"Tai Yi", 101009, 81}, {"Yang Jian", 101321, 63}, {"Huang Feihu", 101567, 72}, {"Zhou Wang", 108160, 55}, {"Li Jing", 102456, 84}, {"Tu Xingsun", 102245, 65}, }; int n = sizeof(students) / sizeof(Student); TreeNode* root = new TreeNode(students[0]); queue<TreeNode*> q; q.push(root); for (int i = 1; i < n; i += 2) { TreeNode* parent = q.front(); q.pop(); parent->left = new TreeNode(students[i]); q.push(parent->left); if (i + 1 < n) { parent->right = new TreeNode(students[i + 1]); q.push(parent->right); } } cout << "Post-order traversal: " << endl; postOrder(root); cout << "Pre-order traversal: " << endl; preOrder(root); cout << "Level-order traversal: " << endl; levelOrder(root); return 0; }

可以使用'\n'代替endl，这样可以避免刷新缓存区的操作，提高输出效率。修改后的代码如下：

#include <iostream>
#include <queue>
#include <stack>
#include <string>

using namespace std;

struct Student {
    string name;
    int number;
    int score;
};

struct TreeNode {
    Student data;
    TreeNode* left;
    TreeNode* right;
    TreeNode(Student s) : data(s), left(nullptr), right(nullptr) {}
};

void postOrder(TreeNode* root) {
    if (!root) return;
    postOrder(root->left);
    postOrder(root->right);
    cout << root->data.name << " " << root->data.number << " " << root->data.score << '\n';
}

void preOrder(TreeNode* root) {
    stack<TreeNode*> s;
    s.push(root);
    while (!s.empty()) {
        TreeNode* node = s.top();
        s.pop();
        if (node) {
            cout << node->data.name << " " << node->data.number << " " << node->data.score << '\n';
            s.push(node->right);
            s.push(node->left);
        }
    }
}

void levelOrder(TreeNode* root) {
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        TreeNode* node = q.front();
        q.pop();
        if (node) {
            cout << node->data.name << " " << node->data.number << " " << node->data.score << '\n';
            q.push(node->left);
            q.push(node->right);
        }
    }
}

int main() {
    Student students[] = {
        {"Lei Zhenzi", 101401, 82},
        {"Jiang Ziya", 100032, 90},
        {"Ne Zha", 101674, 70},
        {"Shen Gongbao", 101982, 87},
        {"Jiu Weihu", 107431, 75},
        {"Tian Zun", 100001, 98},
        {"Tai Yi", 101009, 81},
        {"Yang Jian", 101321, 63},
        {"Huang Feihu", 101567, 72},
        {"Zhou Wang", 108160, 55},
        {"Li Jing", 102456, 84},
        {"Tu Xingsun", 102245, 65},
    };
    int n = sizeof(students) / sizeof(Student);

    TreeNode* root = new TreeNode(students[0]);
    queue<TreeNode*> q;
    q.push(root);
    for (int i = 1; i < n; i += 2) {
        TreeNode* parent = q.front();
        q.pop();
        parent->left = new TreeNode(students[i]);
        q.push(parent->left);
        if (i + 1 < n) {
            parent->right = new TreeNode(students[i + 1]);
            q.push(parent->right);
        }
    }

    cout << "Post-order traversal: \n";
    postOrder(root);

    cout << "Pre-order traversal: \n";
    preOrder(root);

    cout << "Level-order traversal: \n";
    levelOrder(root);

    return 0;
}