1.采用201933110041进行数据测试 编写二差树
2.要求
具体代码,实验结果截图,代码注释
详细代码如下,望采纳。
//二叉树结构
class TreeNode {
public:
TreeNode(int val_) :val(val_), left(NULL), right(NULL) {}
TreeNode* left;
TreeNode* right;
int val;
};
//前序遍历递归
void headPrint1(TreeNode* root) {
if (root == nullptr) return;
cout<<root->val<<endl;
headPrint1(root->left);
headPrint1(root->right);
}
//前序遍历非递归
void headPrint2(TreeNode* root) {
if (root == nullptr) return;
stack<TreeNode*> stk;
TreeNode* node = root;
stk.push(root);
while (!stk.empty()) {
node = stk.top();
stk.pop();
cout << node->val<<endl;
if (node->right != nullptr) stk.push(node->right);
if (node->left != nullptr) stk.push(node->left);
}
}
//后序遍历递归
void tailPrint1(TreeNode* root) {
if (root == nullptr) return;
tailPrint1(root->left);
tailPrint1(root->right);
cout << root->val << endl;
}
//后序遍历非递归
void tailPrint1(TreeNode* root) {
vector<int> res;
if (root == nullptr) return;
stack<TreeNode*> stk;
TreeNode* prev = nullptr;
while (root != nullptr || !stk.empty()) {
while (root != nullptr) {
stk.emplace(root);
root = root->left;
}
root = stk.top();
stk.pop();
if (root->right == nullptr || root->right == prev) {
cout<<root->val<<endl;
prev = root;
root = nullptr;
}
else {
stk.emplace(root);
root = root->right;
}
}
}
给定一个二叉树,返回它的前中后序遍历。
示例:输入[1,2,3,4,-1,5,6,-1,7,-1,-1,8,-1] ,-1代表无节点(null)
输出:二叉树的前序、中序、后序遍历序列。
完整实现代码(包括定义数据结构、建树等)如下:
#include<iostream>
#include<string>
#include<vector>
#include<cstring>
#include<stack>
#include<map>
#include<queue>
using namespace std;
struct TreeNode{
int val;
TreeNode *left,*right;
TreeNode(int x):val(x),left(NULL),right(NULL){}
};
//---建树
TreeNode* creaTree(vector<int> x){
queue<TreeNode*>nodequeue;
TreeNode *root=new TreeNode(x[0]);
nodequeue.push(root);
TreeNode *p=root;
int i=1;
while(true){
p=nodequeue.front();
nodequeue.pop();
if(i>=x.size())break;
if(x[i]!=-1){
p->left=new TreeNode(x[i]);
nodequeue.push(p->left);
}
i ++;
if(i>=x.size())break;
if(x[i]!=-1){
p->right=new TreeNode(x[i]);
nodequeue.push(p->right);
}
i++;
}
return root;
}
class Solution {
public:
//-------------前序递归---
vector<int>res1;
vector<int> recpreorderTraversal(TreeNode* root){
if(root){
res1.push_back(root->val);
recpreorderTraversal(root->left);
recpreorderTraversal(root->right);
}
return res1;
}
//------------前序非递归
vector<int> preorderTraversal(TreeNode* root){
vector<int>res;
TreeNode *p=root;
stack<TreeNode*>s;
while(p || !s.empty()){
while(p){
res.push_back(p->val);
s.push(p);
p=p->left;
}
if(!s.empty()){
p=s.top();
s.pop();
p=p->right;
}
}
return res;
}
//---------中序非递归
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*>s;
vector<int>res;
//map<TreeNode*,bool>instack;
TreeNode *p=root;
//s.push(root);
while(p!=nullptr || !s.empty()){
while(p!=nullptr){
s.push(p);
p=p->left;
}
if(!s.empty()){
p=s.top();
s.pop();
//cout<<p->val;
res.push_back(p->val);
p=p->right;
}
}
return res;
}
//--------------中序递归---------
vector<int>res2;
vector<int> recinorderTraversal(TreeNode* root) {
if (root != nullptr) {
recinorderTraversal(root->left);
res2.push_back(root->val);
recinorderTraversal(root->right);
}
return res2;
}
//-------------后序递归
vector<int>res3;
vector<int> recpostorderTraversal(TreeNode* root) {
if(root){
recpostorderTraversal(root->left);
recpostorderTraversal(root->right);
res3.push_back(root->val);
}
return res3;
}
//---------------后序非递归
vector<int> postorderTraversal(TreeNode* root){
vector<int>res;
stack<TreeNode*>s;
TreeNode *p,*pre= nullptr;
s.push(root);
while(!s.empty()){
p=s.top();
if((!p->left && !p->right) || ((pre!=nullptr) &&(pre==p->left ||pre==p->right))){
res.push_back(p->val);
s.pop();
pre=p;
}
else{
if(p->right)
s.push(p->right);
if(p->left)
s.push(p->left);
}
}
return res;
}
};
int main(){
vector<int>x={1,2,3,4,-1,5,6,-1,7,-1,-1,8,-1};
//cout<<x.size();
TreeNode *root=creaTree(x);
vector<int>res=Solution().postorderTraversal(root);
for(int i=0;i<res.size();i++) {
cout << res[i] << " ";
}
return 0;
}