PAT甲级1099

最新推荐文章于 2022-09-13 23:30:00 发布

原创最新推荐文章于 2022-09-13 23:30:00 发布 · 692 阅读

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本文介绍了一种根据给定的二叉树结构和一组整数键来构建唯一二叉搜索树的方法，并通过示例详细解释了如何进行层级遍历输出。输入包括节点数量、子节点信息及整数键，输出为构建后的树的层级遍历序列。

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1099. Build A Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
Both the left and right subtrees must also be binary search trees.
Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format "left_index right_index", provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.
Sample Input:
```
9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42
```
Sample Output:
```
58 25 82 11 38 67 45 73 42
```

#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 100;
struct node
{
	int data;
	int l, r;
	node():l(-1),r(-1){}
}nodes[maxn];
int index = 0; vector<int> v;
void inorder(int root)
{
	if (root == -1)
		return;
	inorder(nodes[root].l);
	nodes[root].data = v[index++];
	inorder(nodes[root].r);
}
void levelorder(int root)
{
	queue<int> Q;
	if (root != -1)
	{
		Q.push(root);
	}
	bool first = true;
	while (!Q.empty())
	{
		int t = Q.front();
		Q.pop();
		if (first) 
		{
			printf("%d", nodes[t].data); first = false;
		}
		else printf(" %d", nodes[t].data);
		if (nodes[t].l != -1) Q.push(nodes[t].l);
		if (nodes[t].r != -1) Q.push(nodes[t].r);
	}
}
int main()
{
	int N;
	scanf("%d", &N);
	for (int i = 0; i < N; i++)
	{
		scanf("%d %d", &nodes[i].l, &nodes[i].r);
	}
	int t;
	for (int i = 0; i < N; i++)
	{
		scanf("%d", &t);
		v.push_back(t);
	}
	sort(v.begin(), v.end());
	inorder(0);
	levelorder(0);
	return 0;
}