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优化以下代码#include <stdio.h> #include <stdlib.h> typedef struct TreeNode { char data; struct TreeNode* left; struct TreeNode* right; } TreeNode; TreeNode* createTree() { char ch; TreeNode* root; scanf("%c", &ch); if (ch == '#') { return NULL; } root = (TreeNode)malloc(sizeof(TreeNode)); root->data = ch; root->left = createTree(); root->right = createTree(); return root; } void digui(TreeNode root) { if (root == NULL) { return; } digui(root->left); printf("%c ", root->data); digui(root->right); } typedef struct StackNode { TreeNode* tree; struct StackNode* next; } StackNode; typedef struct Stack { StackNode* top; int size; } Stack; Stack* createStack() { Stack* stack = (Stack)malloc(sizeof(Stack)); stack->top = NULL; stack->size = 0; return stack; } void push(Stack stack, TreeNode* tree) { StackNode* node; node = (StackNode)malloc(sizeof(StackNode)); node->tree = tree; node->next = stack->top; stack->top = node; stack->size++; } TreeNode pop(Stack* stack) { TreeNode* tree; StackNode* temp; if (stack->size == 0) { return NULL; } tree = stack->top->tree; temp = stack->top; stack->top = stack->top->next; stack->size--; free(temp); return tree; } void feidigui(TreeNode* root) { Stack* stack; TreeNode* p; stack = createStack(); p = root; while (p != NULL || stack->size != 0) { while (p != NULL) { push(stack, p); p = p->left; } if (stack->size != 0) { p = pop(stack); printf("%c ", p->data); p = p->right; } } } int getHeight(TreeNode* root) { int leftHeight,rightHeight,max; if (root == NULL) { return 0; } leftHeight = getHeight(root->left); rightHeight = getHeight(root->right); max=leftHeight>rightHeight?leftHeight:rightHeight; return max+1; }

stack->top = stack->top->next; stack->size--; free(temp); return tree; } // 非递归遍历二叉树 void feidigui(TreeNode *root) { Stack *stack = createStack(); TreeNode *p = root; while (p...

void inOrderS(TreeNode* root) { stack<TreeNode> S; TreeNode p = root; while (p || !S.empty()){ if(p->left){ S.push(p); p = p->left; } else{ cout << S.top()->val; p = S.top()->right; S.pop(); } } }

a. 如果p不为空，则将p压入栈S中，并将p移动到它的左子节点。 b. 如果p为空，则弹出栈顶元素，输出其值，并将p移动到弹出元素的右子节点。 3. 如果栈S为空并且p也为空，则遍历结束。这段代码实现了二叉树的...

代码分析#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; }

这段代码实现了二叉树的三种遍历方式：后序遍历、前序遍历和层序遍历。它首先定义了两个结构体：Student 和 TreeNode。Student 是一个学生的结构体，包括姓名、学号和分数；TreeNode 是二叉树的结构体，包括...

这段Java代码简易版怎么写String []symbols = chanshengshi.split("->")[1].split(""); for (int i = symbols.length - 1; i >= 0; i--) { stack.push(symbols[i]); }

stack.push(symbols[i]); } 解释一下每行代码的作用： 1. 定义一个字符串变量 chanshengshi，初始化为 "hello->world"。 2. 使用 split() 方法将字符串 chanshengshi 按照 "->" 进行分割，得到一个字符...

#include<stdio.h> #include<stdlib.h> #define StackSize 50 typedef struct { int a[StackSize]; int top; }SeqStack; void InitStack(SeqStack S) { S->top = -1; } int Push(SeqStack S, int x) { if(S->top == StackSize-1) { printf("栈满\n"); } else { S->top++; S->a[S->top] = x; } return 0; } int Pop(SeqStack S, int x) { if(S->top == -1) { printf("空栈\n"); } else { x = S->a[S->top]; S->top--; } return 0; } int GetTop(SeqStack S, int x) { if(S->top == -1) { printf("栈为空"); } else { x = S->a[S->top]; printf("%d",*x); } return 0; } int TTB(SeqStack S, int x) { while(x > 0) { if(x%2 == 0) { Push(&S,0); } else { Push(&S,1); } x = x/2; } while(S.top > -1) { GetTop(&S, &x); S.top--; } return 0; } int main() { SeqStack S; int j; InitStack(&S); scanf_s("%d", &j); TTB(S,j); system("pause"); return 0; }以上代码的思路

S->a[S->top] = x; } return 0; } int Pop(SeqStack *S, int *x) { if(S->top == -1) { printf("空栈\n"); } else { *x = S->a[S->top]; S->top--; } return 0; } int GetTop(SeqStack *S, int *x) ...

#include <stdio.h> #include<malloc.h> #include<math.h> #define STACK_INIT_SIZE 100 #define STACKINCREMENT 10 #define OK 1 #define FALSE 0 typedef int Statue; typedef int SElement; //typedef struct { // int a; //}SElement; SElement N; SElement e; typedef struct { SElement* base; SElement* top; int stacksize; }SqStack; SqStack s; Statue InitStack(SqStack s) { s->base = (SElement)malloc(STACK_INIT_SIZE * sizeof(SElement)); if (!s->base) exit(OVERFLOW); s->top = s->base; s->stacksize = STACK_INIT_SIZE; return OK; } Statue StackEmpty(SqStack s) { if (s.top == -1) return OK; else return FALSE; } Statue Push(SqStack* s, SElement e) { if (s->top - s->base >= s->stacksize) { s->base = (SElement)malloc(s->base, (s->stacksize + STACKINCREMENT) sizeof(SElement)); if (!s->base) exit(OVERFLOW); s->top = s->base + s->stacksize; s->stacksize += STACKINCREMENT; } s->top++ = e; return OK; } Statue Pop(SqStack s, SElement* e) { if (s->top == s->base) return FALSE; e = --s->top; return OK; } void conversion() { //SElement N; //SElement e; InitStack(&s); scanf_s("%d", N); while (&N) { Push(&s, N % 8); N = N / 8; } while (!StackEmpty) { Pop(&s, e); printf("%d", e); } } int main() { InitStack(&s); conversion(); return 0; }

而不是 malloc(s->base, (s->stacksize + STACKINCREMENT) * sizeof(SElement));，这里应该只传入一个参数，即需要分配的内存大小。 2. 在 StackEmpty 函数中，应该使用 s.top == s.base 判断栈是否为空，...

#include<iostream> using namespace std; #include <stack> // 定义树节点结构体 typedef struct TreeNode { char val;//数据域 TreeNode* left;//左孩子 TreeNode* right;//右孩子 }Tree, TreeNode; void CreateTree(Tree& T) { char x; cin >> x; if (x =='') { T = NULL; return; } else { T = new TreeNode; T->val = x; CreateTree(T->left); CreateTree(T->right); } } // 先序遍历二叉树 void preOrderTraversal(TreeNode* root) { if (root == NULL) return; cout << root->val << endl; preOrderTraversal(root->left); preOrderTraversal(root->right); } // 中序遍历二叉树 void inOrderTraversal(TreeNode* root) { if (root == NULL) return; inOrderTraversal(root->left); cout << root->val << endl; inOrderTraversal(root->right); } void inOrderS(TreeNode* root) { stack<TreeNode> S; TreeNode p = root; while (p || !S.empty()){ if(p->left){ S.push(p); p = p->left; } else{ cout << S.top()->val; p = S.top()->right; S.pop(); } } } // 后序遍历二叉树 void postOrderTraversal(TreeNode* root) { if (root == NULL) return; postOrderTraversal(root->left); postOrderTraversal(root->right); cout << root->val <<endl;} int main() { TreeNode* root = NULL; cout << "请输入二叉树的先序遍历序列，以*表示空节点" << endl; CreateTree(root); stack<int> S; //cout << "先序遍历结果为："<< endl; //preOrderTraversal(root); cout << endl << "中序遍历结果为：" << endl; inOrderS(root); //cout << endl << "后序遍历结果为：" << endl; //postOrderTraversal(root); cout << endl; return 0; } 纠错

1. 第一行代码应该是 #include <iostream> 而非 #include<iostream>。 2. 定义树节点结构体时，结构体名称应为 TreeNode 而非 *Tree。 3. 在 CreateTree 函数中，需要加上 new 关键字来动态分配内存。...

#include <stdio.h> #include<iostream> #include<stdlib.h> #include<stdio.h> #define MAXSIZE 20 using namespace std; struct BiTreeNode//二叉树结点定义 { BiTreeNode* LChild;//左孩子指针域 int data; BiTreeNode* RChild;//右孩子指针域 }; struct Stack//栈的定义 { int base;//栈底指针 int top;//栈顶指针 BiTreeNode BTNS[MAXSIZE];//二叉树结点数组 int stackSize;//栈可用的最大容量 }; void InitStack(Stack& S)//初始化栈 { S = (Stack)malloc(sizeof(Stack)); S->top = S->base = 0; S->stackSize = MAXSIZE; } bool StackEmpty(Stack& S)//判断栈是否为空 { if (S->base == S->top) { return true; } else { return false; } } bool StackFull(Stack& S)//判断栈是否已满 { if (S->top - S->base == S->stackSize) { //栈已满 return true; } else { //栈不满 return false; } } void Push(Stack& S, BiTreeNode& T)//元素入栈 { if (StackFull(S) == true) { //如果栈已满, 则直接返回 return; } S->BTNS[S->top].data = T->data; S->BTNS[S->top].LChild = T->LChild; S->BTNS[S->top].RChild = T->RChild; S->top++; } BiTreeNode* Pop(Stack& S)//元素出栈 { if (StackEmpty(S) == true) { return NULL; } S->top--; return &(S->BTNS[S->top]); } // void CreateBiTree(BiTreeNode& T)//以先序序列创建二叉树 { char ch; cin >> ch; if (ch != '#') { T = (BiTreeNode)malloc(sizeof(BiTreeNode)); T->data = ch; CreateBiTree(T->LChild); CreateBiTree(T->RChild); } else { T = NULL; } } void InOrderTraverse(Stack& S, BiTreeNode& T)//中序遍历二叉树的非递归算法(※) { InitStack(S);//初始化栈 BiTreeNode p = T; BiTreeNode* q; while (p || !StackEmpty(S)) { if (p) { Push(S, p); p = p->LChild; } else { q = Pop(S);//出栈元素指针保存在q中 putchar(q->data); cout << " "; p = q->RChild; } } } int main() { Stack* S; BiTreeNode* T; CreateBiTree(T); InOrderTraverse(S, T); return 0; }请帮我把代码优化一下

#include <stack> using namespace std; struct BiTreeNode { // 二叉树结点定义 BiTreeNode* LChild; // 左孩子指针域 char data; BiTreeNode* RChild; // 右孩子指针域 }; void CreateBiTree(BiTreeNode*& T)...

#include <stdio.h> #include <stdlib.h> #define STACK_INIT_SIZE 100 #define STACK_INCREMENT 10 typedef struct { int base; int top; int stack_size; } Stack; void InitStack(Stack S) { S->base = (int )malloc(STACK_INIT_SIZE * sizeof(int)); //为栈申请初始空间 if (!S->base) { exit(0); } S->top = S->base; S->stack_size = STACK_INIT_SIZE; } void Push(Stack S, int e) { if (S->top - S->base >= S->stack_size) { //栈空间已满，需要扩容 S->base = (int )realloc(S->base, (S->stack_size + STACK_INCREMENT) * sizeof(int)); if (!S->base) { exit(0); } S->top = S->base + S->stack_size; S->stack_size += STACK_INCREMENT; } (S->top) = e; //入栈 S->top++; } int Pop(Stack S, int e) { if (S->top == S->base) { //栈为空，出栈失败 return 0; } S->top--; e = *(S->top); //出栈 return 1; } int main() { int n, e; Stack S; InitStack(&S); printf("请输入一个十进制整数："); scanf("%d", &n); while (n) { Push(&S, n % 2); // 将除2取余的结果入栈 n /= 2; } printf("转换为二进制的结果是："); while (Pop(&S, &e)) { // 依次出栈，输出结果 printf("%d", e); } printf("\n"); return 0; }解释一下这这程序每一个代码的作用，或者在整个程序中的效果

if (S->top - S->base >= S->stack_size) { // 栈空间已满，需要扩容 S->base = (int *)realloc(S->base, (S->stack_size + STACK_INCREMENT) * sizeof(int)); if (!S->base) { exit(0); } S->top = S->base +...

Please write a simply code in python, the requirement is as followed: The Tower of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The following figure shows the initial state of the Tower of Hanoi with 5 disks.The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: 1. Only one disk can be moved at a time. 2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. 3. No disk may be placed on top of a smaller disk. Assume that initially all the disks are placed on rod A. Write a non-recursive Python function to print out the steps to move all the disks from rod A to rod C via rod B (Hint: a recursive algorithm can be converted into a non-recursive algorithm using stack). The header of the function is: def HanoiTower(n) Here n represents the number of disks. example: when n = 3, the program will output: A -> C 换行 A -> B 换行 C -> B 换行 A -> C 换行 B -> A 换行 B -> C 换行 A -> C

stack.append((n, 'A', 'B', 'C')) while len(stack) > 0: n, from_rod, aux_rod, to_rod = stack.pop() if n == 1: print(from_rod, "->", to_rod) else: stack.append((n-1, aux_rod, from_rod, to_rod...

Please write a simply code in python, the requirement is as followed: The Tower of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The following figure shows the initial state of the Tower of Hanoi with 5 disks.The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: 1. Only one disk can be moved at a time. 2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. 3. No disk may be placed on top of a smaller disk. Assume that initially all the disks are placed on rod A. Write a non-recursive Python function to print out the steps to move all the disks from rod A to rod C via rod B (Hint: a recursive algorithm can be converted into a non-recursive algorithm using stack). The header of the function is: def HanoiTower(n) Here n represents the number of disks. example: when n = 3, the program will output: A -> C next line A -> B next line C -> B next line A -> C next line B -> A next line B -> C next line A -> C 4. simple enough

while len(stack) > 0: n, fr, to, via = stack.pop() if n == 1: print(fr, '->', to) else: stack.append((n-1, via, to, fr)) stack.append((1, fr, to, via)) stack.append((n-1, fr, via, to)) # 调用...

stack.push

stack.push是栈的一个操作，用于将元素压入栈顶。在给定的代码中，stack是一个MyStack类的实例，通过调用push方法可以将元素添加到栈中。具体实现是将元素添加到数组的栈顶位置，并更新栈顶指针。例如，如果要将元素...

public List<Integer> preorderTraversal(TreeNode root) { List<Integer> result = new ArrayList<>(); if (root == null) return result; Stack<TreeNode> stack = new Stack<>(); stack.push(root); while (!stack.isEmpty()) { TreeNode node = stack.pop(); result.add(node.val); if (node.right != null) stack.push(node.right); if (node.left != null) stack.push(node.left); } return result; } }

这段代码是一个二叉树的前序遍历算法实现。它使用了栈来辅助遍历。首先，将根节点压入栈中。然后，循环执行以下操作：弹出栈顶节点，将其值加入结果列表中，然后将其右子节点（如果存在）压入栈中，最后将其左子节点...

给我写出完整的可运行的代码public class Solution { public String removeDuplicateLetters(String s) { int[] tem = new int[26]; for (int i = 0; i < s.length(); i++) { char tems = s.charAt(i); tem[tems - 'a']++; } boolean[] visited = new boolean[26]; Stack<Character> stack = new Stack<>(); for (int i = 0; i < s.length(); i++) { char get = s.charAt(i); while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) { visited[stack.peek() - 'a'] = false; stack.pop(); } if (visited[get - 'a']) { tem[get - 'a']--; continue; } visited[get - 'a'] = true; stack.push(get); tem[get - 'a']--; } StringBuilder sBuilder = new StringBuilder(); while (!stack.isEmpty()) { sBuilder.append(stack.pop()); } return sBuilder.reverse().toString(); }的 }

visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) { visited[stack.peek() - 'a'] = false; stack.pop(); } if (visited[get - 'a']) { tem[get - 'a']--; continue; } visited[get - 'a'] = true; ...

aStack.push(pointer->rightchild());

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这段Java代码简易版怎么写String []symbols = chanshengshi.split("->")[1].split(""); for (int i = symbols.length - 1; i >= 0; i--) { stack.push(symbols[i]); }

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void inOrderS(TreeNode* root) { stack<TreeNode> S; TreeNode p = root; while (p || !S.empty()){ if(p->left){ S.push(p); p = p->left; } else{ cout << S.top()->val; p = S.top()->right; S.pop(); } } }