PAT日志 1086

顽强的小白

1086 Tree Traversals Again （25 分）

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.

图片自己脑部

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

题目解析

用栈的形式来表示二叉树，这里就会发现一个有趣的现象，所有入栈的操作序列就是先序遍历，而所有出栈的序列就是中序遍历。发现这个现象后就和上一道二叉树的题目一样了，这次是给出中序和先序求后序遍历。

代码实现

#include <cstdio> 
#include <algorithm>
#include <stack>
#include <cstring>
using namespace std;
const int maxn=35;
int n;
int in[maxn];
int pre[maxn];
int post[maxn];
struct node{
 int data;
 node* l;
 node* r;
};
node* create(int inL,int inR,int preL,int preR){
 if(inL>inR) return NULL;
 node* root=new node;
 root->data=pre[preL];
 int k;
 for(k=inL;k<=inR;++k){
  if(in[k]==pre[preL])
   break;
 }
 int left=k-inL;
 root->l=create(inL,k-1,preL+1,preL+left);
 root->r=create(k+1,inR,preL+left+1,preR);
 return root;
}
int cnt=0;
void postOrder(node* root){
 if(root==NULL) return;
 postOrder(root->l);
 postOrder(root->r);
 printf("%d",root->data);
 cnt++;
 if(cnt<n) printf(" ");
 
}
int main(){
 stack<int> s;
 scanf("%d",&n);
 char mk[6];
 int j=0,k=0,x;
 for(int i=0;i<n*2;++i){
  scanf("%s",mk);
  if(strcmp(mk,"Push")==0){
   scanf("%d",&x);
   pre[j++]=x;
   s.push(x);
  }else{
   in[k++]=s.top();
   s.pop();
  }
 }
 node* root=create(0,n-1,0,n-1);
 postOrder(root);
 return 0;
}