1086 Tree Traversals Again (25 分)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

入栈push顺序为先序，出栈pop顺序为中序，根据先序中序序列还原二叉树，最终输出后序序列

#include<bits/stdc++.h>
using namespace std;

int n,x,in[40],pre[40],inI=0,preI=0;
stack<int> s;

struct node{
	int data;
	node* lchild;
	node* rchild;
};

//申请内存 为所有字段赋初值 
node* newNode(int v){
	node* Node=new node;
	Node->data=v;
	Node->lchild=Node->rchild=NULL;
	return Node;
}

node* create(int preL,int preR,int inL,int inR){
	if(preL>preR) return NULL;
	node* root=newNode(pre[preL]);
	int k;//中序中根结点位置 
	for(k=inL;k<=inR;k++){
		if(in[k]==pre[preL]) break;
	}
	int lnum=k-inL;//左子树长度
	root->lchild=create(preL+1,preL+lnum,inL,k-1); 
	root->rchild=create(preL+lnum+1,preR,k+1,inR); 
	return root;
}

bool first=true;
void postOrder(node* root){
	if(root==NULL) return;
	postOrder(root->lchild);
	postOrder(root->rchild);
	if(!first) cout<<" ";
	else first=false;
	cout<<root->data;
}

int main(){
//	freopen("in.txt","r",stdin);
	string op;
	cin>>n;
	int T=2*n;
	while(T--){
		cin>>op;
		if(op=="Push"){
			cin>>x;
			s.push(x);
			pre[inI++]=x;
		}else{
			in[preI++]=s.top();
			s.pop();
		}
	} 
	postOrder(create(0,n-1,0,n-1));
	return 0;
}

注意，pat的输入可能不是很标准，根据‘ ’判断最后一组数据过不了。猜想也可能是最后一组数据最后没有换行，导致getline()读不进最后一个pop.总之，输入cin可以很好地过滤各种空格换行，不要投机取巧，标准输入输出求稳

bug代码

#include<bits/stdc++.h>
using namespace std;

int n,x,in[40],pre[40],inI=0,preI=0;
stack<int> s;

void print(int a[]){
	for(int i=0;i<n;i++) cout<<a[i]<<" ";
	cout<<endl;
}

struct node{
	int data;
	node* lchild;
	node* rchild;
};

//申请内存 为所有字段赋初值 
node* newNode(int v){
	node* Node=new node;
	Node->data=v;
	Node->lchild=Node->rchild=NULL;
	return Node;
}

node* create(int preL,int preR,int inL,int inR){
	if(preL>preR) return NULL;
	node* root=newNode(pre[preL]);
	int k;//中序中根结点位置 
	for(k=inL;k<=inR;k++){
		if(in[k]==pre[preL]) break;
	}
	int lnum=k-inL;//左子树长度
	root->lchild=create(preL+1,preL+lnum,inL,k-1); 
	root->rchild=create(preL+lnum+1,preR,k+1,inR); 
	return root;
}

bool first=true;
void postOrder(node* root){
	if(root==NULL) return;
	postOrder(root->lchild);
	postOrder(root->rchild);
	if(!first) cout<<" ";
	else first=false;
	cout<<root->data;
}

int main(){
//	freopen("in.txt","r",stdin);
	string op;
	cin>>n;
	cin.get();
	int T=2*n;
	while(T--){
		getline(cin,op);
		if(op.find(' ')!=string::npos){
			x=op[op.length()-1]-'0';
			s.push(x);
			pre[inI++]=x;
		}else{
			in[preI++]=s.top();
			s.pop();
		}
	} 
	
	node* root=create(0,n-1,0,n-1);
	postOrder(root);
	
	return 0;
}

来源：CSDN

作者：树叶子_

链接：https://blog.csdn.net/hza419763578/article/details/100527729

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