浙大PAT 1020. Tree Traversals (25)

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

/*
 * 题意：已知二叉树的后中序输出层序遍历的结果。
 * 通过后中序建立树 bfs输出层序结果即可。 
*/
#include "iostream"
#include "queue"
using namespace std;
struct Node {
	int val;
	Node* l;
	Node* r;
};
typedef Node* Tree;
Tree t;
Tree buildTree(int* post,int* in,int n) {
	if (n <= 0)
		return NULL;
	int i;
	for (i = 0; post[n-1]!=in[i]; i++);
	Tree t = (Tree)malloc(sizeof(struct Node));
	t->val = post[n - 1];
	t->l = buildTree(post,in,i);
	t->r = buildTree(post+i,in+i+1,n-i-1);
	return t;
}
void bfs() {
	queue<Tree>q;
	q.push(t);
	bool flag = 1;
	while (!q.empty()) {
		Tree p = q.front();
		if (flag) {
			cout << p->val;
			flag = false;
		}
		else
			cout << " " << p->val;
		if (p->l != NULL)
			q.push(p->l);
		if (p->r != NULL)
			q.push(p->r);
		q.pop();
	}
}
int main() {
	int n;
	int post[31];
	int in[31];
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> post[i];
	for (int i = 0; i < n; i++)
		cin >> in[i];
	t = buildTree( post, in, n);
	bfs();
	cout << endl;
	return 0;
}