PAT A1020【Tree Traversals】(25 分)

1020 Tree Traversals (25 分)

Problem Description

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

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

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

中规中矩的一道题，根据中序和后序构建树，然后进行层次遍历:

#include<iostream>
#include<queue>
using namespace std;

//https://pintia.cn/problem-sets/994805342720868352/problems/994805485033603072

const int maxn = 32;
int postorder[maxn];
int inorder[maxn];

struct Node{
	int data;
	Node *lchild, *rchild;
};

Node *create_tree(int inL, int inR, int postL, int postR)
{
	if(inL > inR)
		return NULL;
	Node *root = new Node;
	root->data = postorder[postR]
	int k;
	for(k=inL; k<=inR; k++){
		if(inorder[k] == postorder[postR])
			break;
	}
	int numLeft = k-inL;
	root->lchild = create_tree(inL, k-1, postL, postL+numLeft-1);
	root->rchild = create_tree(k+1, inR, postL+numLeft, postR-1);
	return root;
}

void layerorder(Node *root)
{
	queue<Node*> q;
	q.push(root);
	bool space_flag = false;
	while(!q.empty()){
		Node *node = q.front();
		q.pop();
		if(space_flag)
			cout<<" ";
		space_flag = true;
		cout<<node->data;
		if(node->lchild)
			q.push(node->lchild);
		if(node->rchild)
			q.push(node->rchild);
	}
}

void in(Node *root)
{
	if(root == NULL)
		return;
	in(root->lchild);
	cout<<root->data<<" ";
	in(root->rchild);
}

int main()
{
	int N;
	cin>>N;
	for(int i=1; i<=N; i++)
		cin>>postorder[i];
	for(int i=1; i<=N; i++)
		cin>>inorder[i];
	Node *root = create_tree(1, N, 1, N);
	layerorder(root);
	return 0;
 }