PAT 甲级 1086 Tree Traversals Again

1086 Tree Traversals Again （25 point(s)）

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

经验总结：

这一题，实际上题目已经告诉你二叉树的前序序列以及中序序列，然后求后序序列就可以啦~  理解题目意思，就没什么难度了~中间可以不用建树，直接根据前序与中序得出后序序列，这样应该可以节省一点时间~

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;
const int maxn=40;
int in[maxn],post[maxn],pre[maxn],n,p,postlen=0;
void convert(int inL,int inR,int preL,int preR)
{
	if(preL>preR)
		return ;
	int k;
	for(k=inL;k<=inR;++k)
		if(in[k]==pre[preL])
			break;
	convert(inL,k-1,preL+1,preL+k-inL);
	convert(k+1,inR,preL+k-inL+1,preR);
	post[postlen++]=pre[preL];
}
int main()
{
	char s[7];
	int t,inlen=0,prelen=0;
	stack<int> q;
	scanf("%d",&n);
	for(int i=0;i<n+n;++i)
	{
		scanf("%s",s);
		if(strcmp(s,"Push")==0)
		{
			scanf("%d",&t);
			q.push(t);
			pre[prelen++]=t;
		}
		else
		{
			in[inlen++]=q.top();
			q.pop();
		}
	}
	convert(0,inlen-1,0,prelen-1);
	for(int i=0;i<postlen;++i)
		printf("%d%c",post[i],i<postlen-1?' ':'\n');
	return 0;
}