树的应用:Tree Traversals Again

题目来自:pta题目集（无邀请码打不开）
题目如下:
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
输入样例:

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

输出样例:

3 4 2 6 5 1

分析过程:

通过观察可以发现，元素入栈的顺序为该树的先序遍历，出栈 顺序为中序遍历，
那么这个问题便可转化为已知先序和中序遍历求后序遍历

代码如下:

#include <cstdio>
#include <time.h>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
char s[6],tt[6]="Push",r[6]="Pop";
int a[33],b[33],aaa=0,bbb=0,c[33],ccc=0;
struct node{int data;node*pre;node*left;node*right;
};
node* newnode(int key){node*t=(node*)malloc(sizeof(node));t->data=key;t->pre=NULL;t->left=NULL;t->right=NULL;return t;
}//创建新节点
void buildtree(int l,int r,int &t,node*&tree){int flag=-1;for(int i=l;i<=r;i++){if(b[i]==a[t]) {flag=i;break;}}if (flag==-1) return;tree=newnode(b[flag]);t++;if(flag>l) buildtree(l,flag-1,t,tree->left);if(flag<r) buildtree(flag+1,r,t,tree->right);
}
void post(node*tree){if(tree){post(tree->left);post(tree->right);c[ccc++]=tree->data;}
}
int main()
{   int n,m,k,cnt=0,t=0;scanf("%d",&n);stack<int>kkk;while(1){scanf("%s",s);if(strcmp(s,tt)==0){scanf("%d",&k);a[aaa++]=k;kkk.push(k);}else{b[bbb++]=kkk.top();kkk.pop();cnt++;}if(cnt==n) break;}node*root;buildtree(0,n-1,t,root);post(root);for(int i=0;i<n;i++){printf("%d",c[i]);if(i!=n-1) printf(" ");}return 0;
}

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