Homework New
This commit is contained in:
e2hang
97
Exercise/Homework4/Q1.cpp
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97
Exercise/Homework4/Q1.cpp
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#include <iostream>
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#include <vector>
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using namespace std;
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template <class T>
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struct Node {
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T element;
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Node<T>* left;
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Node<T>* right;
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Node(const T& e) : element(e), left(nullptr), right(nullptr) {}
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};
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template <class T>
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class BinTree {
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public:
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Node<T>* root;
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int height;
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vector<T> arr;
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BinTree() : root(nullptr), height(0) {}
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BinTree(const vector<T>& arrs) : arr(arrs), height(0) {}
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void test() {
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for (auto x : arr) {
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cout << x << " ";
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}
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}
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Node<T>* returnroot() {
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return this->root;
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}
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Node<T>* build(int& index) {
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if (index >= arr.size() || arr[index] == 0) {
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index++;
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return nullptr;
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}
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Node<T>* node = new Node<T>(arr[index]);
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index++;
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node->left = build(index);
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//cout << "build left ";
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node->right = build(index);
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//cout << "build right ";
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return node;
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}
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void buildTree() {
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int index = 0;
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this->root = build(index);
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}
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//<2F><><EFBFBD><EFBFBD>
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void preorder(Node<T>* node) {
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if (node == nullptr) return;
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cout << node->element << " ";
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preorder(node->left);
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preorder(node->right);
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}
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void inorder(Node<T>* node) {
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if (node == nullptr) return;
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inorder(node->left);
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cout << node->element << " ";
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inorder(node->right);
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}
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void postorder(Node<T>* node) {
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if (node == nullptr) return;
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postorder(node->left);
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postorder(node->right);
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cout << node->element << " ";
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}
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};
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int main() {
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int x = 0;
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vector<int> arr;
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int tmp;
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Node<int>* parent;
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while (cin >> tmp) {
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arr.push_back(tmp);
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}
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BinTree<int> tree(arr);
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tree.buildTree();
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//tree.test();
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auto r = tree.root;
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tree.preorder(r);
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cout << endl;
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tree.inorder(r);
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cout << endl;
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tree.postorder(r);
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cout << endl;
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return 0;
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}
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121
Exercise/Homework4/Q2.cpp
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121
Exercise/Homework4/Q2.cpp
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#include <iostream>
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#include <vector>
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#include<string>
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#include <sstream>
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using namespace std;
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template <class T>
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struct Node {
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T element;
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Node<T>* left;
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Node<T>* right;
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Node<T>* parent;
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Node(const T& e, Node<T>* p) : element(e), parent(p), left(nullptr), right(nullptr) {}
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};
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template <class T>
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class BinTree {
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public:
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Node<T>* root;
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int height;
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vector<T> arr;
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BinTree() : root(nullptr), height(0) {}
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BinTree(const vector<T>& arrs) : arr(arrs), height(0) {}
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void test() {
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for (auto x : arr) {
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cout << x << " ";
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}
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}
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Node<T>* returnroot() {
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return this->root;
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}
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Node<T>* build(int& index, Node<T>* parent) {
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if (index >= arr.size() || arr[index] == 0) {
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index++;
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return nullptr;
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}
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Node<T>* node = new Node<T>(arr[index], parent);
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index++;
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node->left = build(index, node);
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node->right = build(index, node);
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return node;
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}
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void buildTree() {
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int index = 0;
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this->root = build(index, nullptr);
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}
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//<2F><><EFBFBD><EFBFBD>
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void preorder(Node<T>* node) {
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if (node == nullptr) return;
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cout << node->element << " ";
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preorder(node->left);
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preorder(node->right);
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}
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void inorder(Node<T>* node) {
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if (node == nullptr) return;
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inorder(node->left);
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cout << node->element << ", parent";
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//cout << ( node->parent == nullptr ? 0 : node->parent->element) << endl;
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inorder(node->right);
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}
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void postorder(Node<T>* node) {
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if (node == nullptr) return;
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postorder(node->left);
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postorder(node->right);
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cout << node->element << " ";
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}
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Node<T>* inorder2(Node<T>* node, const T& x) {
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if (node == nullptr) return nullptr;
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if (node->element == x) {
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return node;
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}
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Node<T>* leftResult = inorder2(node->left, x);
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if (leftResult != nullptr) {
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return leftResult;
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}
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Node<T>* rightResult = inorder2(node->right, x);
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return rightResult;
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}
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void findparent(const T& e) {
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auto x = inorder2(this->root, e);
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if (x == nullptr || x->parent == nullptr) cout << 0 << endl;
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else cout << x->parent->element << endl;
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}
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};
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int main() {
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int x = 0;
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vector<int> arr;
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string line;
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getline(cin, line);
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stringstream ss(line);
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int num;
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while (ss >> num) {
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arr.push_back(num);
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}
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BinTree<int> tree(arr);
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tree.buildTree();
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int m;
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cin >> m;
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for (int i = 0; i < m; i++) {
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int t;
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cin >> t;
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tree.findparent(t);
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}
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return 0;
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}
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181
Exercise/Homework4/Q3.cpp
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181
Exercise/Homework4/Q3.cpp
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#include <iostream>
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#include <vector>
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#include<string>
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using namespace std;
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template <class T>
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struct Node {
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T element;
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Node<T>* left;
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Node<T>* right;
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Node<T>* parent;
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Node(const T& e, Node<T>* p) : element(e), parent(p), left(nullptr), right(nullptr) {}
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};
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template <class T>
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class BinTree {
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public:
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Node<T>* root;
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int height;
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vector<T> arr;
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BinTree() : root(nullptr), height(0) {}
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BinTree(const vector<T>& arrs) : arr(arrs), height(0) {}
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void test() {
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for (auto x : arr) {
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cout << x << " ";
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}
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}
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Node<T>* returnroot() {
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return this->root;
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}
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Node<T>* build(int& index, Node<T>* parent) {
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if (index >= arr.size() || arr[index] == 0) {
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index++;
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return nullptr;
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}
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Node<T>* node = new Node<T>(arr[index], parent);
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index++;
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node->left = build(index, node);
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node->right = build(index, node);
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return node;
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}
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void buildTree() {
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int index = 0;
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this->root = build(index, nullptr);
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}
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Node<T>* create(const string& post, const string& in,
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int ps, int pe, int is, int ie, bool& ok) {
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if (ps > pe || is > ie) {
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return nullptr;
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}
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T root = post[pe];
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int pos = -1;
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for (int i = is; i <= ie; i++) {
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if (in[i] == root) {
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pos = i;
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break;
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}
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}
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if (pos == -1) {
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ok = false;
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return nullptr;
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}
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Node<T>* node = new Node<T>(root, nullptr);
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int len = pos - is;
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node->left = create(post, in, ps, ps + len - 1, is, pos - 1, ok);
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node->right = create(post, in, ps + len, pe - 1, pos + 1, ie, ok);
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return node;
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}
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int getHeight(Node<T>* node) {
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if (node == nullptr) return 0;
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int lh = getHeight(node->left);
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int rh = getHeight(node->right);
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if (lh > rh) return lh + 1;
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else return rh + 1;
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}
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string preorderStr(Node<T>* node) {
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if (node == nullptr) return "";
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string res = "";
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res += node->element;
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res += preorderStr(node->left);
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res += preorderStr(node->right);
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return res;
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}
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//<2F><><EFBFBD><EFBFBD>
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void preorder(Node<T>* node) {
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if (node == nullptr) return;
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cout << node->element << " ";
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preorder(node->left);
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preorder(node->right);
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}
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void inorder(Node<T>* node) {
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if (node == nullptr) return;
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inorder(node->left);
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cout << node->element << ", parent";
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//cout << ( node->parent == nullptr ? 0 : node->parent->element) << endl;
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inorder(node->right);
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}
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void postorder(Node<T>* node) {
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if (node == nullptr) return;
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postorder(node->left);
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postorder(node->right);
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cout << node->element << " ";
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}
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Node<T>* inorder2(Node<T>* node, const T& x) {
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if (node == nullptr) return nullptr;
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if (node->element == x) {
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return node;
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}
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Node<T>* leftResult = inorder2(node->left, x);
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if (leftResult != nullptr) {
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return leftResult;
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}
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Node<T>* rightResult = inorder2(node->right, x);
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return rightResult;
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}
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void findparent(const T& e) {
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auto x = inorder2(this->root, e);
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if (x == nullptr || x->parent == nullptr) cout << 0 << endl;
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else cout << x->parent->element << endl;
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}
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};
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bool check(const string& post, const string& in) {
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if (post.length() != in.length()) return false;
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int cnt[26] = {0};
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for (int i = 0; i < post.length(); i++) {
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cnt[post[i] - 'A']++;
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}
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for (int i = 0; i < in.length(); i++) {
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cnt[in[i] - 'A']--;
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if (cnt[in[i] - 'A'] < 0) return false;
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}
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return true;
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}
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int main() {
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string post, in;
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while (getline(cin, post) && getline(cin, in)) {
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if (post.empty() || in.empty()) break;
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if (!check(post, in)) {
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cout << "INVALID" << endl;
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continue;
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}
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BinTree<char> tree;
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bool ok = true;
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tree.root = tree.create(post, in, 0, post.length() - 1, 0, in.length() - 1, ok);
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if (!ok) {
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cout << "INVALID" << endl;
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} else {
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tree.height = tree.getHeight(tree.root) - 1;
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cout << tree.height << endl;
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string pre = tree.preorderStr(tree.root);
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cout << pre << endl;
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}
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}
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return 0;
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}
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93
Exercise/Homework4/readme.md
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93
Exercise/Homework4/readme.md
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@@ -0,0 +1,93 @@
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Q1
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```
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通过带空指针信息的先根序列（亦称先序序列）创建二叉树，并进行先根（先序）、中根（中序）、后根（后序）遍历。二叉树结点数据域值为不等于0的整数（可能是正数也可能是负数），空指针用0表示，例如1 5 8 0 0 0 6 0 0表示如下图的二叉树。
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PA567.jpg
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输入格式:
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输入为一组用空格间隔的整数，表示带空指针信息的二叉树先根序列。其中空指针信息用0表示。二叉树结点个数不超过150000，高度不超过6000。输入数据保证二叉树各结点数据值互不相等。
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输出格式:
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输出为3行整数，每个整数后一个空格。第1行为该二叉树的先根序列，第2行为中根序列，第3行为后根序列。
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输入样例:
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1 5 8 0 0 0 6 0 0
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输出样例:
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1 5 8 6
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8 5 1 6
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8 5 6 1
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代码长度限制
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16 KB
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时间限制
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200 ms
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内存限制
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20 MB
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栈限制
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8192 KB
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```
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Q2
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```
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编写程序在二叉树中查找给定结点及父结点。二叉树结点的数据域值不等于0的整数。
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输入格式:
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输入第1行为一组用空格间隔的整数，表示带空指针信息的二叉树先根序列，其中空指针用0表示。例如1 5 8 0 0 0 6 0 0表示如下图的二叉树。第2行为整数m，表示查询个数。接下来m行，每行为一个不等于0的整数K，表示要查找的结点的数据值。m不超过100，二叉树结点个数不超过150000，高度不超过6000。输入数据保证二叉树各结点数据值互不相等。
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PA567.jpg
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输出格式:
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输出为m行，每行1个整数，表示被查找结点K的父结点数据值，若二叉树中无结点K或结点K无父结点，则输出0。
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输入样例:
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1 5 8 0 0 0 6 0 0
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3
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8
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1
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6
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输出样例:
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5
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0
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1
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代码长度限制
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16 KB
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时间限制
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300 ms
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内存限制
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20 MB
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栈限制
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131000 KB
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```
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Q3
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```
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给定非空二叉树的中根序列和后根序列，请编写程序创建该二叉树，计算其高度和先根序列；如给定的中根和后根序列不合法，则亦能识别。
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输入格式:
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输入包含多组数据（不超过10组），每组为两行字符串，第一行表示某二叉树的后根序列，第二行表示其中根序列。结点的值均为A-Z的大写字母，故二叉树结点个数不超过26，且保证输入的两个序列都是结点的全排列，但不一定是合法的中根和后根序列。输入保证不是空二叉树。
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输出格式:
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对于每组数据，如果输入的序列不合法（不是同一棵树的中根序列和后根序列），则输出INVALID；若输入序列合法，输出为两行，第一行为一个整数，表示该二叉树的高度，第二行为一个字符串，表示该二叉树的先根序列。
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输入样例1:
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CEFDBHGA
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CBEDFAGH
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CBEDFAGH
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CEFDBHGA
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BCA
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CAB
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输出样例1:
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3
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ABCDEFGH
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INVALID
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INVALID
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代码长度限制
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16 KB
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时间限制
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50 ms
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内存限制
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64 MB
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栈限制
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||||
8192 KB
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```
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Reference in New Issue
Block a user