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
#include
#include
#include
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 q;
scanf("%d",&n);
for(int i=0;i