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

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

PAT大考在即，这是16年秋季的一道考题。

题意：有一棵二叉树，现在给你它的前序遍历，后序遍历，问你它的中序遍历是否唯一。如果唯一，请输出Yes和它的结果。不唯一，则输出No和它的其中一组结果。

解题思路：

0.先通过前序找到左子树的顶点，然后确定左子树范围，递归将左子树push到vector0中，再放入根节点。然后同理处理右子树。

1.通过后序找到右子树顶点，然后确定左子树范围，递归将左子树push到vector1中，再放入根节点。然后同理处理右子树。

3.比较vector1和vector2来确定唯一性，输出vector1。

上源码：注释里有不少解释，应该能讲清楚。

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
using namespace std;

typedef long long int ll;

vector<int> pre;
vector<int> post;
vector<int> ansa;
vector<int> ansb;

void pre_post(int st1,int ed1,int st2,int ed2){
    int index;
    //如果找到一个后序的节点与前序第二个节点一样(即左子树的根)
    for(int i=st2; i<=ed2; i++)
        if(post[i]==pre[st1+1]){
            index = i;
            break;
        }
    //左子树就一个节点
    if(st2==index)
        ansa.push_back(post[st2]);

    else if(index>st2)
        pre_post(st1+1,st1+index-st2+1,st2,index);//递归处理左子树
	
    ansa.push_back(pre[st1]);  //放入中间节点
    //右子树就一个节点
    if(index+1==ed2-1)
        ansa.push_back(post[index+1]);
	
    else if(ed2-1 > index+1)
        pre_post(st1+index-st2+2,ed1,index+1,ed2-1);//递归处理右子树
}
//与上基本同理 就是从右子树开始判断
void post_pre(int st1,int ed1,int st2,int ed2){
    int index;
    for(int i=st1; i<=ed1; i++)
        if(pre[i]==post[ed2-1]){
            index = i;
            break;
        }
		
	//左子树就一个节点
    if(st1+1==index-1)
        ansb.push_back(pre[st1+1]);
	
    else if(st1+1<index-1)
        post_pre(st1+1,index-1,st2,st2+index-st1-2);//递归处理左子树
	
    ansb.push_back(pre[st1]); //放入中间节点
	//右子树就一个节点
    if(index==ed1)
        ansb.push_back(pre[index]);
	
    else if(index<ed1)
        post_pre(index,ed1,st2+index-st1-1,ed2-1);//递归处理右子树
}



int main(){

    int N;
    scanf("%d",&N);
    int num;
    for(int i=1; i<=N; i++){
        scanf("%d",&num);
        pre.push_back(num);
    }
    for(int i=1; i<=N; i++){
        scanf("%d",&num);
        post.push_back(num);
    }
    if(N==1){
        printf("Yes\n");
        printf("%d\n",pre[0]);
        return 0;
    }
    pre_post(0,N-1,0,N-1);
    post_pre(0,N-1,0,N-1);

    if(ansa==ansb)
        printf("Yes\n");
    else
        printf("No\n");

    cout << ansa[0];
    for(int i=1; i<ansa.size(); i++)
        cout << " " << ansa[i];
    cout << endl;

    return 0;
}