1119. Pre- and Post-order Traversals (30)

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最新推荐文章于 2024-05-07 12:26:38 发布

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分类专栏： PAT A_LEVEL 文章标签： pat-数据结构

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本文介绍如何利用给定的预序和后序遍历序列来确定二叉树的中序遍历序列。即使对应的二叉树可能不是唯一的，文章也提供了一种方法来构建并输出至少一种可能的中序遍历序列。

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1119. Pre- and Post-order Traversals (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Special

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first printf in a line "Yes" if the tree is unique, or "No" if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input 1:

7
1 2 3 4 6 7 5
2 6 7 4 5 3 1

Sample Output 1:

Yes
2 1 6 4 7 3 5

Sample Input 2:

4
1 2 3 4
2 4 3 1

Sample Output 2:

No
2 1 3 4
#include<iostream>
#include<string>
using namespace std;
struct tree{
	int element;
	tree* left;
	tree* right;
};
int issure=1;
void buildtree(int *s1,int *s2,tree *t,int N){
	int i;
	for(i=0;i<N;i++){
		if(*(s1+i)==*(s2+N-2))
		break;
	}
	t->element=*s1;
//	cout<<s1.size()<<"*"<<endl;
	if(N==2)
	issure=0;
	//cout<<s1<<" "<<s2<<"qizhogs1[0]="<<s1[0]<<endl;
	if(i-1>0){
		t->left=new tree;
	buildtree(s1+1,s2,t->left,i-1);
	}
	else
	t->left=NULL;
	if(N-i>0){
		t->right=new tree;
	buildtree(s1+i,s2+i-1,t->right,N-i);
	}
	else
	t->right=NULL;
}
int fir=0;
void inorder(tree *t){
	if(t->left!=NULL)
	inorder(t->left);
	if(fir==0)
	fir=1;
	else
	cout<<" ";
		cout<<t->element;
	if(t->right!=NULL)
	inorder(t->right);
}
int main(){
	int N,i,j,k;
	cin>>N;
	int *a=new int[N];
	int *b=new int[N]; 
	int c;
	string s1,s2;
	for(i=0;i<N;i++) {
		cin>>c;
		a[i]=c;
	}
//	cout<<s1;
	for(i=0;i<N;i++) {
		cin>>c;
		b[i]=c;
	}
//	cout<<s2;
	tree* t;
	t=new tree;
	buildtree(a,b,t,N);
	if(issure)
	cout<<"Yes"<<endl;
	else
	cout<<"No"<<endl;
	inorder(t);
	cout<<endl;
} 

感想：做法
1.最开始当然是建树了，找规律，前序遍历与后序遍历有一个特点，前序遍历的第二个元素是左子树的根，后序遍历的倒数第二个元素是右子树的根。既然已经知道了右子树的根，我们在前序遍历里找到这个右子树的根，我们又知道了树的总根，又知道左子树的根，这样就可以把前序遍历和后序遍历的序列分为3个部分（根，左子树，右子树），然后再分别对左子树和右子树递归建树。这样就建立出来了
2.不要用字符串存树，因为这样就不能存大于10的树了
3.最后要加一个换行符才不会格式错误