Pre- and Post-order Traversals（前序后序求中序（不唯一）

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本文探讨了基于前序和后序遍历序列确定二叉树中序遍历序列的算法，分析了序列对应二叉树可能存在的唯一性问题，并提供了一个具体的实现示例，展示了如何从输入序列中计算出中序遍历序列。

给出前序遍历和后序遍历求一个中序遍历并判断是否唯一，如果孩子全在左/右子树上并且没有兄弟节点就会出现不唯一的情况

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<bits/stdc++.h>
#include<set>
#include<vector>
#include<map>
#include<queue>
using namespace std;

int pre[10005];
int pos[10005];
vector<int>ans;
bool flag=true;

void intra(int prel,int prer,int posl,int posr) {
	if(prel==prer) {
		ans.push_back(pre[prel]);
		return ;
	}
	if(pre[prel]==pos[posr]) {
		int i=prel+1;
		while(pre[i]!=pos[posr-1])i++;
		if(i-prel<=1)
			flag=false;
		else {
			intra(prel+1,i-1,posl,posl+(i-prel-1)-1);
		}
		ans.push_back(pos[posr]);
		intra(i,prer,posl+(i-prel-1),posr-1);
	}
}
int main() {

	int n;
	cin>>n;
	for (int i =1; i <=n; i++) 
		scanf("%d", &pre[i]);
	for (int i = 1; i <=n; i++)
		scanf("%d", &pos[i]);
		
	
	intra(1,n,1,n);
	printf("%s\n%d", flag == true ? "Yes" : "No", ans[0]);
	for (int i = 1; i < ans.size(); i++)
		printf(" %d", ans[i]);
	printf("\n");


	return 0;
}