PAT甲级——A1020 Tree Traversals

题目

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 (≤), 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

Solution:

已知后序遍历和中序遍历输出层序遍历
这里是使用重构二叉树来求解的
当然也可以不用重构二叉树，直接根据数值的数组位置即可

Code

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
struct Node
{
	int v;
	Node *l, *r;
	Node(int a = -1) :v(a), l(nullptr), r(nullptr) {}
};
vector<int>pos, in, level;
int n;
Node* reCreateTree(int inL, int inR, int posL, int posR)
{
	if (inL > inR)
		return nullptr;
	Node* root = new Node(pos[posR]);
	int k, m;
	for (int i = inL; i <= inR; ++i)
	{
		if (in[i] == pos[posR])
		{
			k = i;
			break;
		}
	}
	m = k - inL;
	root->l = reCreateTree(inL, k - 1, posL, posL + m - 1);
	root->r = reCreateTree(k + 1, inR, posL + m, posR - 1);
	return root;
}
void levelOrder(Node* root)
{
	if (root == nullptr)
		return;
	queue<Node*>q;
	q.push(root);
	while (!q.empty())
	{
		Node* p = q.front();
		q.pop();
		level.push_back(p->v);
		if (p->l != nullptr)
			q.push(p->l);
		if (p->r != nullptr)
			q.push(p->r);
	}
}
int main()
{
	cin >> n;
	pos.resize(n);
	in.resize(n);
	for (int i = 0; i < n; ++i)
		cin >> pos[i];
	for (int i = 0; i < n; ++i)
		cin >> in[i];
	Node* root = reCreateTree(0, n - 1, 0, n - 1);
	levelOrder(root);
	for (int i = 0; i < n; ++i)
		cout << (i == 0 ? "" : " ") << level[i];
	return 0;
}