【PTA-A】1020 Tree Traversals (25 分)（二叉树、BFS）

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

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

Sample Output:

4 1 6 3 5 7 2

#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 50;
struct Node {
	int data;
	Node* lchild;
	Node* rchild;
};
int pre[MAXN], in[MAXN], post[MAXN];
int n;

Node* create(int postl, int postr, int inl, int inr) {
	if (postl > postr) return NULL;
	Node* root = new Node;
	root->data = post[postr];
	int k;
	for (k = inl; k <= inr; k++) {
		if (in[k] == post[postr])
			break;
	}
	int numl = k - inl;
	root->lchild = create(postl, postl + numl - 1, inl, k - 1);
	root->rchild = create(postl + numl, postr - 1, k + 1, inr);
	return root;
}

int num = 0;
void BFS(Node* root) {
	queue<Node*> q;
	q.push(root);
	while (!q.empty()) {
		Node* now = q.front();
		q.pop();
		cout << now->data;
		num++;
		if (num < n)cout << " ";
		if (now->lchild != NULL)q.push(now->lchild);
		if (now->rchild != NULL)q.push(now->rchild);
	}
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> post[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> in[i];
	}
	Node* root = create(0, n - 1, 0, n - 1);
	BFS(root);
	return 0;
}