A1119 Pre- and Post-order Traversals [前序后序转中序]

前序和后序之所以不确定一棵树的原因就是当只有一棵子树的时候，不确定是左子树还是右子树。找前序的第二个结点在后续中位置，左边就是左子树。递归建树
要是到最后一个位置还没找出根节点，那么区分不了左右，标志位标记false

#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
using namespace std;
const int maxn = 50;
struct node
{
	int data;
	node* lchild;
	node* rchild;
};
int pre[maxn], post[maxn];
int n;
bool isUnique = true;
node* create(int prel, int prer, int postl, int postr)
{
	if (prel > prer)
		return NULL;
	node* root = new node;
	root->data = pre[prel];
	int i, numLeft = 0;
	for (i = postl;i <postr; i++)
	{
		numLeft++;
		if (post[i] == pre[prel + 1])
			break;
	}
	if (i == postr - 1)  
		isUnique = false;
	root->lchild = create(prel + 1, prel + numLeft, postl,i);
	root->rchild = create(prel + numLeft+1, prer, i + 1, postr - 1);
	return root;
}
int flag = 1;
void inorder(node* root)
{

	if (root == NULL)
		return;
	inorder(root->lchild);
	if (flag)
	{
		cout << root->data;
		flag = 0;
	}
	else
		cout << " " << root->data;
	inorder(root->rchild);
}
int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> pre[i];
	}
	for (int i = 0; i < n; i++)
	{
		cin >> post[i];
	}
	node* root = create(0, n - 1, 0, n - 1);	
	if (isUnique)
	{
		cout << "Yes" << endl;
		inorder(root);
		cout << endl;
	}
	else
	{
		cout << "No" << endl;
		inorder(root);
		cout << endl;
	}

}