PAT甲级 -- 1020 Tree Traversals (25 分)

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最新推荐文章于 2025-01-24 22:50:08 发布

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博客围绕二叉树展开，已知二叉树所有键为不同正整数，给出后序和中序遍历序列，要求输出对应二叉树的层序遍历序列。还给出了具体思路，即后序中序转层序，在转前序过程增加结点下标变量，递归输出先序并排序得到层序。

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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

25分：

思路：

1. 后序中序转层序，与转前序相似，在转前序的过程中，增加变量index，表示当前结点在二叉树的下标（从1开始）

2. 进行一次输出先序的递归，把结点的index和key值放入vector，然后进行排序，即可按照层序输出

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;



struct node
{
	int index;
	int v;
};

bool cmp(node a, node b) {
	return a.index < b.index;
}
int n;
vector<int> postOrder, inOrder;
vector<node> v;

void LayOrder(int inl, int inr, int postr,int index){
	if (inl > inr) return;
	int i = inl;
	while (inl < inr && postOrder[postr] != inOrder[i]) i++;
	node a;
	a.index = index;
	a.v = postOrder[postr];
	v.push_back(a);
	LayOrder(inl, i-1, postr-1-inr+i, 2 * index);
	LayOrder(i+1, inr, postr-1, 2 * index + 1);
}


int main(){
	scanf("%d", &n);
	postOrder.resize(n),inOrder.resize(n);
	for (int i = 0; i < n; i++) scanf("%d", &postOrder[i]);
	for (int i = 0; i < n; i++) scanf("%d", &inOrder[i]);
	LayOrder(0,n-1,n-1,0);
	sort(v.begin(), v.end(), cmp);
	for (int i = 0; i < v.size(); i++) {
		if (i != 0) cout << " ";
		cout << v[i].v;
	}
	return 0;
}