解释这行代码if(k<a. length()-1) preorder(a. substr(k+1,a. length()-k-1),b. substr( k,b. length()-k-1));

js-leetcode题解之144-binary-tree-preorder-traversal.js

javascript js_leetcode题解之144-binary-tree-preorder-traversal.js

数据结构实验代码先序+中序序列构造树.rar

这个"数据结构实验代码先序+中序序列构造树.rar"文件显然包含了一个关于如何通过先序和中序遍历序列来构建二叉树的实验代码。在这里，我们将深入探讨这两个关键的二叉树遍历方法以及它们在构建二叉树中的应用。 **...

preorder-2.cpp

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; }

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 = ...

代码改进，用printf代替cout。#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; }

好的，这里是使用 printf 替换 cout 后的代码： #include <cstdio> #include <queue> #include <stack> #include <string> using namespace std; struct Student { string name; int number; int score; }; ...

#include<stdio.h> #include<iostream.h> #include<stdlib.h> typedef char ElementType; typedef struct TNode { ElementType Data; struct TNode left, right; }TNode, BiTree; void InOrderTraverse(BiTree T) { if(T) { InOrderTraverse(T->left); cout<<T->Data; InOrderTraverse(T->right); } } void Preorder_Traversal(BiTree T) { if(!T)return; cout<<T->Data; Preorder_Traversal(T->left); Preorder_Traversal(T->right); } void Postorder_Traversal(BiTree T) { if(!T)return; Postorder_Traversal(T->left); Postorder_Traversal(T->right); cout<<T->Data; } int Depth(BiTree T) { int m,n; if(T==NULL) return 0; else { m=Depth(T->left); n=Depth(T->right); if(m>n) return(m+1); else return(n+1); } } int NodeCount(BiTree T) { if(T==NULL) return 0; else return NodeCount(T->left)+NodeCount(T->right)+1; } int CountLeaf(BiTree T, int* count) { if (T!=NULL&&(T->left)== NULL&&(T->right)==NULL) { (count)++; } if (T!= NULL) { CountLeaf(T->left, count); CountLeaf(T->reght, count); } return count; }

这段代码有很多问题，我来一一指出： 1. #include<iostream.h> 应该改为 #include<iostream>，因为 <iostream.h> 是旧版 C++ 标准库，现在已经被弃用。 2. 在 CountLeaf 函数中，T 是一个指向指针的指针...

#include <bits/stdc++.h> using namespace std; void getpost(string preorder,string inorder) //根据先序和中序求后序 { int n = preorder.length(); //n为每次遍历数目 if(n > 0) { char root = preorder[0]; //根结点为先序遍历的第一个 int i = inorder.find(root); //中序遍历中根结点的所在下标 getpost(preorder.substr(1,i),inorder.substr(0,i)); //左子树 getpost(preorder.substr(i+1),inorder.substr(i+1)); //右子树 cout << root; } } int main() { string preorder,inorder; //先序遍历和中序遍历 while(cin >> preorder >> inorder) { getpost(preorder,inorder); cout << endl; } return 0; }请用c语言转换这个代码

以下是将给定的C++代码转换为等效的C语言代码： #include <stdio.h> #include <string.h> void getpost(char* preorder, char* inorder) { int n = strlen(preorder); if(n > 0) { char root = preorder[0...

import java.util.Arrays; public class Create { public TreeNode CreateTree(int[] preorder,int[] inorder) { if (preorder.length == 0 || inorder.length == 0) { return null; //前序中序长度都为0 返回空 } TreeNode root = new TreeNode(preorder[0]);//从前序中获取根节点。 int rootIndex = 0; //获取根节点在中序遍历中的位置 for (int i = 0; i < inorder.length; i++) { if (root.val == inorder[i]) { rootIndex = i; } } int[] Inorderleft = Arrays.copyOfRange(inorder, 0, rootIndex);//把中序遍历从根节点拆开左右两部分这是左部分 int[] Inorderright = Arrays.copyOfRange(inorder, rootIndex + 1, inorder.length);//把中序遍历从根节点拆开右部分 int[] Preorderleft= Arrays.copyOfRange(preorder,0,rootIndex);//把先序遍历已根节点 int[] Preorderright=Arrays.copyOfRange(preorder,rootIndex+1,preorder.length); root.left=CreateTree(Preorderleft,Inorderleft); root.right=CreateTree(Preorderright,Inorderright); return root; } }

这段代码定义了一个名为Create的类，并且包含了一个名为CreateTree的方法。该方法用于根据给定的前序遍历数组preorder和中序遍历数组inorder创建一棵二叉树。方法首先进行了一些判断，如果前序遍历数组或...

s1.pop(); s2.push(curr); if (curr->left) s1.push(curr->left); if (curr->right) s1.push(curr->right); } while (!s2.empty()) { TreeNode* curr = s2.top(); s2.pop(); result.push_back(curr->val); } return result; } int main() { string preorder = "ABDGCEF"; // 先序序列 string inorder = "DGBAECF"; // 中序序列 TreeNode* root = buildTree(preorder, inorder); // 构建二叉树 vector<char> preorderResult = preorderTraversal(root); // 前序非递归遍历 vector<char> inorderResult = inorderTraversal(root); // 中序非递归遍历 vector<char> postorderResult = postorderTraversal(root); // 后序非递归遍历 // 输出三种遍历序列的结果 cout << "前序非递归遍历: "; for (size_t i = 0; i < preorderResult.size(); i++) { cout << preorderResult[i] << " "; } cout << endl; cout << "中序非递归遍历: "; for (size_t i = 0; i < inorderResult.size(); i++) { cout << inorderResult[i] << " "; } cout << endl; cout << "后序非递归遍历: "; for (size_t i = 0; i < postorderResult.size(); i++) { cout << postorderResult[i] << " "; } cout << endl; return 0; }

这是一个用 C++ 实现的二叉树构建和三种非递归遍历的代码。在这个代码中，先序遍历、中序遍历和后序遍历都采用非递归的方式实现。这个代码的具体实现过程如下： 1. 首先，根据给定的先序序列和中序序列构建二叉树。...

#include <stdio.h> #include <stdlib.h> typedef struct TreeNode { int val; struct TreeNode left; struct TreeNode right; } TreeNode; TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize) { if (preorderSize == 0) { return NULL; } int i; for (i = 0; i < inorderSize; i++) { if (inorder[i] == preorder[0]) { break; } } TreeNode* root = (TreeNode)malloc(sizeof(TreeNode)); root->val = preorder[0]; root->left = buildTree(preorder + 1, i, inorder, i); root->right = buildTree(preorder + i + 1, preorderSize - i - 1, inorder + i + 1, inorderSize - i - 1); return root; } void inorderTraversal(TreeNode root) { if (root == NULL) { return; } inorderTraversal(root->left); printf("%d ", root->val); inorderTraversal(root->right); } int main() { int preorder[] = {3, 9, 20, 15, 7}; int inorder[] = {9, 3, 15, 20, 7}; int len = sizeof(preorder) / sizeof(int); TreeNode* root = buildTree(preorder, len, inorder, len); inorderTraversal(root); return 0; }

这段代码使用 C 语言实现了根据前序遍历序列和中序遍历序列构造二叉树，并输出中序遍历结果。 c #include <stdio.h> // 导入标准输入输出库 #include <stdlib.h> // 导入标准库 // 定义二叉树结构体 typedef ...

#include<stdio.h> #include<stdlib.h> int a[31], b[31]; typedef struct treeNode* tree; struct treeNode { int data; tree left, right; }; tree creat() { tree p = (tree)malloc(sizeof(tree)); p->left = p->right = NULL; return p; } tree find(int z1, int x1, int x2)//寻找根节点 { int i; tree head = creat(); head->data = a[z1]; for (i = x1; i <= x2; i++) { if (a[z1] == b[i]) { if(i!=x1) //有左支树 head->left=find(z1-(x2-i)-1,x1,i-1); if(i!=x2) //有右支树 head->right=find(z1-1,i+1,x2); } } return head; } void PreOrder(tree head) { if (head) { printf(" %d", head->data); PreOrder(head->left); PreOrder(head->right); } } int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &a[i]); } for (int i = 0; i < n; i++) { scanf("%d", &b[i]); } tree head = find(n - 1, 0, n - 1); printf("Preorder:"); PreOrder(head); return 0; }

这是一段 C 语言的代码，实现了根据二叉树的前序遍历序列和中序遍历序列构建二叉树，并输出二叉树的前序遍历序列。具体实现方法是，先根据前序遍历序列的第一个节点确定根节点，然后在中序遍历序列中找到根节点的...

将一下二叉树代码进行中序线索化实现：#include <stdio.h> #include <stdlib.h> typedef struct TreeNode { int val; struct TreeNode* left; struct TreeNode* right; } TreeNode; TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize) { if (preorderSize == 0 || inorderSize == 0) { return NULL; } TreeNode* root = (TreeNode)malloc(sizeof(TreeNode)); root->val = preorder[0]; root->left = NULL; root->right = NULL; int rootIndex = 0; while (inorder[rootIndex] != root->val) { rootIndex++; } root->left = buildTree(preorder + 1, rootIndex, inorder, rootIndex); root->right = buildTree(preorder + rootIndex + 1, preorderSize - rootIndex - 1, inorder + rootIndex + 1, inorderSize - rootIndex - 1); return root; } int isValid(TreeNode root, int min, int max) { if (root == NULL) { return 1; } if (root->val <= min || root->val >= max) { return 0; } return isValid(root->left, min, root->val) && isValid(root->right, root->val, max); } int main() { int preorder[] = {1, 2, 4, 5, 3, 6, 7}; int inorder[] = {4, 2, 5, 1, 6, 3, 7}; int n = sizeof(preorder) / sizeof(preorder[0]); TreeNode* root = buildTree(preorder, n, inorder, n); if (isValid(root, INT_MIN, INT_MAX)) { printf("二叉树正确\n"); } else { printf("二叉树错误\n"); } return 0; }

最后，我们可以在 main 函数中调用这个函数，将二叉树转化为线索二叉树： ThreadedNode* threadedRoot = createThreadedTree(root); 这个 threadedRoot 就是转化后的线索二叉树的根节点了。

分析代码：class TreeNode: def init(self, val): self.val = val self.left = None self.right = None def build_tree(postorder): if not postorder: return None root = TreeNode(postorder[-1]) i = 0 while i < len(postorder) - 1 and postorder[i] < root.val: i += 1 root.left = build_tree(postorder[:i]) root.right = build_tree(postorder[i:-1]) return root def preorder_traversal(root): if not root: return [] stack, res = [root], [] while stack: node = stack.pop() res.append(node.val) if node.right: stack.append(node.right) if node.left: stack.append(node.left) return res # 测试代码 postorder = [1, 3, 2, 5, 7, 6, 4] root = build_tree(postorder) print(preorder_traversal(root))

这段 Python 代码实现了根据二叉搜索树的后序遍历序列构建二叉搜索树，并输出其先序遍历序列。以下是代码的分析： 1. 定义了一个 TreeNode 类，表示二叉树的节点，包含一个值属性 val 和两个指针属性 left 和...

#include <iostream> #include <vector> #include <sstream> using namespace std; struct TreeNode { string val; TreeNode* left; TreeNode* right; TreeNode(string x) : val(x), left(nullptr), right(nullptr) {} }; TreeNode* buildTreeHelper(const vector<string>& preorder, int& pos, const string& sep) { if (pos >= preorder.size() || preorder[pos] == sep) { ++pos; return nullptr; } string s = preorder[pos]; ++pos; TreeNode* root = new TreeNode(s); root->left = buildTreeHelper(preorder, pos, sep); root->right = buildTreeHelper(preorder, pos, sep); return root; } TreeNode* buildTree(const vector<string>& preorder, const string& sep) { int pos = 0; return buildTreeHelper(preorder, pos, sep); } void preorder1(TreeNode* root) { if (!root) return; cout << root->val << ","; preorder1(root->left); preorder1(root->right); } void inorder(TreeNode* root) { if (!root) return; inorder(root->left); cout << root->val << ","; inorder(root->right); } void postorder(TreeNode* root) { if (!root) return; postorder(root->left); postorder(root->right); cout << root->val << ","; } int main() { string sep; getline(cin, sep); vector<string> preorder; string line; getline(cin, line); stringstream ss(line); string s; while (getline(ss, s, ' ')) { preorder.push_back(s); } TreeNode* root = buildTree(preorder, sep); cout << "Preorder: "; preorder1(root); cout << endl; cout << "Inorder: "; inorder(root); cout << endl; cout << "Postorder: "; postorder(root); cout << endl; return 0; }如何避免在输出序列最后多输出一个逗号

void preorder1(TreeNode* root) { if (!root) return; cout << root->val; if (root->left || root->right) cout << ","; preorder1(root->left); preorder1(root->right); } void inorder(TreeNode* root) ...

#include <stdio.h> #include <stdlib.h> typedef char ElementType; typedef struct BiTNode{ ElementType data; struct BiTNode lchild; struct BiTNode rchild; }BiTNode,*BiTree; BiTree CreatBinTree(); void preorder( BiTree T ); int main() { BiTree T = CreatBinTree(); preorder( T ); return 0; } void preorder( BiTree T ) { if(T) { printf("%c",T->data); preorder(T->lchild); preorder(T->rchild); } } BiTree CreatBinTree() { char ch;BiTree T; scanf("%c",&ch); if(ch=='#') return 1 分 ; T= 1 分 ; T->data=ch; T->lchild= 1 分 ; T->rchild= 1 分 ; return T; }

这段代码实现了二叉树的创建和先序遍历。具体来说，代码中定义了二叉树结构体 BiTNode，并通过 typedef 定义了 BiTree 为指向 BiTNode 的指针类型。其中，BiTNode 中包含了数据元素 data 和指向左右子树的指针 ...

#include<stdio.h> #include<stdlib.h> #include<malloc.h> typedef struct bnode { int data; struct bnode lc; struct bnode rc; }bnode; void creat(bnode proot) { bnode p,q; int k,n,i; proot=NULL;//初始化空树 printf("输入二叉树的节点元素个数n：");//输入数据元素个数 scanf("%d",&n); for(i=0;i<n;i++) { p=(bnode)malloc(sizeof(bnode)); if(p == NULL) {//检查分配内存是否成功 printf("树满\n"); exit(1);} p->lc=NULL; p->rc=NULL; printf("k="); scanf("%d",&k); if(proot==NULL) proot=p; else {q=proot; while(q!=NULL) {if(q->data>k) if(q->lc!=NULL) q=q->lc; else {q->lc=p; q=NULL; } else if(q->rc!=NULL) q=q->rc; else {q->rc=p; q=NULL; } } } } } void visit(bnode proot) { bnode temp=proot->lc; proot->lc=proot->rc; proot->rc=temp; } int preorder(bnode proot)//递归遍历 {if(proot==NULL) print("xxx\n"); return; { printf("%d",proot->data); if(proot->lc!=NULL) preorder(proot->lc); if(proot->rc!=NULL); preorder(proot->rc); } } void print(bnode proot) { int level = -1; //记录是第几层次 int i; if (NULL == proot) return; level++; print(proot->rc); level--; level++; for (i = 0; i < level; i++) printf("\t"); printf("%d\n", proot->data); print(proot->lc); level--; } int main() { bnode *proot=NULL; creat(proot); preorder(proot); print(proot); return 0; }

这段代码实现了二叉树的创建和遍历，并且还包括了一个函数用于翻转二叉树。具体来说，这个程序定义了一个结构体bnode，其中包含了节点值和左右子树的指针。接着，使用creat函数创建了二叉树，该函数通过输入节点元素...

这道题的大部分程序已经完成，题目中的二叉树是根据后序遍历和中序遍历序列建立的，主函数也已经给出，你只需要在适当的位置添加先序遍历二叉树的函数 PreOrder( ) 即可。 #include<stdio.h> #include<stdlib.h> typedef char DataType; typedef struct Node { DataType data; struct Node lchild, rchild; }BiTNode, BinTree; void CreateBinTree_Post_In(BiTNode &t, DataType post[], DataType in[], int s1, int t1, int s2, int t2) { int i; if (s1 <= t1) { t = (BiTNode)malloc(sizeof(BiTNode)); t->data = post[t1]; t->lchild = NULL; t->rchild = NULL; for (i = s2; i <= t2; i++) if (in[i] == post[t1]) break; CreateBinTree_Post_In(t->rchild, post, in, s1 + i-s2, t1-1, i+1, t2); CreateBinTree_Post_In(t->lchild, post, in, s1, s1+i-s2-1, s2, i-1); } } /你的代码段将放在这个位置/ int main() { char post, in; int n; BiTNode t; while (scanf("%d", &n) != EOF) { post = (char)malloc(n sizeof(char)); in = (char)malloc(n sizeof(char)); scanf("%s", post); scanf("%s", in);

void PreOrder(BinTree t) { if (t != NULL) { printf("%c ", t->data); PreOrder(t->lchild); PreOrder(t->rchild); } } int main() { char *post, *in; int n; BiTNode *t; while (scanf("%d", &n) != ...

class Tree{ constructor(value){ this.value=value; this.children=[]; } addChild(value){ this.child.push(new Tree(value)); } inorder(){ if(this.children.length)return[this.value]; let result=[]; for(let child of this.children){ result=result.concat(child.inorder());} return[this.value].concat(result); } preorder(){ let result=[this.value]; for(let child of this.children){ result=result.concat(child.preorder()); } return result; } postorder(){ if(!this.children.length)return[this.value]; let result=[]; for(let child of this.children){ result=result.concat(child.postorder()); } return result.concat([this.value]); } }帮我修改这段代码，并且增加修改和查找

在这个修改后的代码中，我添加了 modifyValue(newValue) 和 findNode(value) 方法。 modifyValue(newValue) 方法可以用于修改树节点的值，你只需要调用该方法并传入新的值即可。 findNode(value) 方法可以...

解释这行代码if(k<a. length()-1) preorder(a. substr(k+1,a. length()-k-1),b. substr( k,b. length()-k-1));

解释这段代码void preorder(string a,string b){ if(b. length() ==1) pre += b; else { int k = a. find(b. substr(b. length()-1,1)); pre += a[k]; if(k>0) preorder(a. substr(0,k),b. substr(0,k)); if(k<a. length()-1) preorder(a. substr(k+1,a. length()-k-1),b. substr( k,b. length()-k-1)); } }

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解释这行代码if(k<a. length()-1) preorder(a. substr(k+1,a. length()-k-1),b. substr( k,b. length()-k-1));

解释这段代码void preorder(string a,string b){ if(b. length() ==1) pre += b; else { int k = a. find(b. substr(b. length()-1,1)); pre += a[k]; if(k>0) preorder(a. substr(0,k),b. substr(0,k)); if(k<a. length()-1) preorder(a. substr(k+1,a. length()-k-1),b. substr( k,b. length()-k-1)); } }

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