03-树2. Tree Traversals Again (25)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose thatwhen 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 Figure1) 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

提交代码

题目的意思是给你inorder的栈操作步骤，让你写出postorder后的序列。

我的思路是先还原原来的树，再进行递归后序。

代码如下。

#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <stack>

using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
typedef struct Node
{
int data;
Node *l;
Node *r;
Node(int data)
{
this->data=data;
}
} *BinTree;

int N=0,count=0,in=0;

BinTree Bulid()
{
char s[10];
int a=0;
BinTree nod=NULL;
if(count<2*N)
{
scanf("%s",s);
if(strcmp(s,"Push") ==0 )
{
scanf("%d",&a);
nod=new Node(a);
count++;
}
else
{
count++;
return NULL;
}
nod->l=Bulid();
nod->r=Bulid();
}
return nod;
}
void post(BinTree bt)
{
if(bt)
{
post(bt->l);
post(bt->r);
in++;
printf("%d",bt->data);
if(in<N) printf(" ");
}
}

int main(int argc, char** argv) {
cin>>N;
Node* root=Bulid();
post(root);

//	cout<<s<<a;

return 0;
}