PAT 甲级 1086 Tree Traversals Again(链表及静态链表实现)

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.

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

解题思路：入栈的序列为先序遍历顺序，出队的顺序为中序遍历顺序，要求输出后序遍历顺序。先对输入的字符串进行处理，将先序，中序序列存在数组中再进行操作。

代码

#include<cstdio>
#include<string>
#include<stack>
#include<iostream>
using namespace std;
stack<int> st;
int in[50],pre[50];
int flag = 0;
struct node{
	int data;
	node* lchild;
	node* rchild;
};
node* create(int preL,int preR,int inL,int inR){
	if(preL > preR){
		return NULL;
	}
	node* root = new node;
	root->data = pre[preL];
	int k;
	for(k = inL;k < inR;k++){
		if(pre[preL] == in[k]){
			break;
		}
	}
	int leftnum = k - inL;
	root->lchild = create(preL+1,preL+leftnum,inL,inL+leftnum - 1);
	root->rchild = create(preL+leftnum+1,preR,inL+leftnum+1,inR);
	return root;
}
void postorder(node* root){
	if(root == NULL)
		return;
	postorder(root->lchild);
	postorder(root->rchild);
	if(flag)
		printf(" ");
	flag = 1;
	printf("%d",root->data);
}

int str_to_num(string s)
{
	int result = 0;
	int a = 1;
	for(int i = s.length() - 1;i>=0;i--){
		result += (s[i] - '0')*a;
		a*=10;
	}
	return result;
}
int main(void)
{
	int N,num,inpos = 0,prepos = 0;
	scanf("%d",&N);
	getchar();
	for(int i = 0;i < 2*N;i++){
		string temp,strnum;
		getline(cin,temp);
		if(temp.length()>3){
			strnum = temp.substr(5);
			num = str_to_num(strnum);
			st.push(num);
			pre[prepos++] = num;
		}
		else{
			num = st.top();
			in[inpos++] = num;
			st.pop();
		}
	}
	node* root = create(0,N - 1,0,N - 1);
	postorder(root);
	return 0;
}

下面用静态链表实现，用结构体数组存储地址，要注意用temp存储index否则index会在递归的过程中改变，导致错误。

#include<cstdio>
#include<string>
#include<stack>
#include<iostream>
using namespace std;
stack<int> st;
int in[50],pre[50];
int flag = 0;
struct node{
	int data;
	int lchild;
	int rchild;
}Node[100];
int index = 0;
int create(int preL,int preR,int inL,int inR){
	if(preL > preR){
		return -1;
	}
	int temp = index;
	index++;
	Node[temp].data = pre[preL];
	int k;
	for(k = inL;k < inR;k++){
		if(pre[preL] == in[k]){
			break;
		}
	}
	int leftnum = k - inL;
	Node[temp].lchild = create(preL+1,preL+leftnum,inL,inL+leftnum - 1);
	Node[temp].rchild = create(preL+leftnum+1,preR,inL+leftnum+1,inR);
	return temp;
}
void postorder(int root){
	if(root == -1)
		return;
	postorder(Node[root].lchild);
	postorder(Node[root].rchild);
	if(flag)
		printf(" ");
	flag = 1;
	printf("%d",Node[root].data);
}

int str_to_num(string s)
{
	int result = 0;
	int a = 1;
	for(int i = s.length() - 1;i>=0;i--){
		result += (s[i] - '0')*a;
		a*=10;
	}
	return result;
}
int main(void)
{
	int N,num,inpos = 0,prepos = 0;
	scanf("%d",&N);
	getchar();
	for(int i = 0;i < 2*N;i++){
		string temp,strnum;
		getline(cin,temp);
		if(temp.length()>3){
			strnum = temp.substr(5);
			num = str_to_num(strnum);
			st.push(num);
			pre[prepos++] = num;
		}
		else{
			num = st.top();
			in[inpos++] = num;
			st.pop();
		}
	}
	int root = create(0,N - 1,0,N - 1);
	postorder(root);
	return 0;
}