PAT甲级1119 Pre- and Post-order Traversals （30 分）

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

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

//题意：给你前序，后序，如果能确定唯一的二叉树，输出Yes，中序输出，如果不唯一，输出No, 取任意一颗树，按中序遍历任意输出

//思路：一完全二叉树为层次遍历为：1 2 3 4 5 6 7，前序为：1 2 4 5 3 6 7，后序为：4 5 2 6 7 3 1 把该样例模拟一遍即为思路；

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
using namespace std;
vector<int>ans;
struct Node
{
    int data;
    Node*l;
    Node*r;
    Node()
    {
        l=NULL;
        r=NULL;
    }
};
int flag=0;

Node *build(int *pre,int *post,int len)
{
    Node *root=new Node;
    if(len==1)
    {
        root->data=pre[0];
        return root;
    }else if(len==0)
    {
        root=NULL;
        return root;
    }
    root->data=pre[0];
    int index=len-2;
    while(pre[1]!=post[index])index--;
    //cout<<*(pre+1)<<" "<<(*post)<<" "<<index+1<<endl;
    //cout<<*(pre+index+2)<<" "<<*(post+index+1)<<" "<<len-2-index<<endl;
    root->l=build(pre+1,post,index+1);
    root->r=build(pre+index+2,post+index+1,len-2-index);

    return root;
}
void check(Node *root)//当一个父节点只有单一一个子节点时，该二叉树不唯一
{
    queue<Node*>q;
    q.push(root);
    while(!q.empty())
    {
        Node *temp=q.front();
        q.pop();
        int cn=0;
        if(temp->l!=NULL)
        {
            q.push(temp->l);
            cn++;
        }
        if(temp->r!=NULL)
        {
            q.push(temp->r);
            cn++;
        }
        if(cn==1)
        {
            flag=1;
            //cout<<temp->data<<endl;
            break;
        }
    }
    if(flag==1)printf("No\n");
    else printf("Yes\n");
}
void inorder(Node *root)
{
    if(root!=NULL)
    {
        inorder(root->l);
        ans.push_back(root->data);
        inorder(root->r);
    }
}
int main()
{
    int pre[35],post[35];
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)scanf("%d",&pre[i]);
    for(int i=0;i<n;i++)scanf("%d",&post[i]);
    Node *root=new Node;
    root=NULL;
    root=build(pre,post,n);
    check(root);
    inorder(root);
    for(int i=0;i<ans.size();i++)
    {
        printf("%d",ans[i]);
        if(i==ans.size()-1)printf("\n");
        else printf(" ");
    }
    return 0;
}