PAT甲级 - 1099 Build A Binary Search Tree (30 分) 重建二叉搜索树

本文介绍了一种基于二叉搜索树(BST)的构建方法,通过递归定义确保左子树的节点键小于根节点,右子树的节点键大于等于根节点,且左右子树也必须是BST。文章详细描述了如何根据树的结构和一组不同的整数键,唯一地填充这些键到树中,以满足BST的定义,并输出树的层次遍历序列。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.

figBST.jpg

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

和A1064很像,中序重建树即可。

//给出树的结构,给出要添加的元素; 按照树的某种遍历是有序的,将元素添加进去
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 150;
struct Node {
	int data;
	int left, right;
}node[maxn];
int origin[maxn];
int n;
int index = 0;

void InOrder(int root) {
	if (root == -1) //到达叶结点
		return;
	InOrder(node[root].left);
	node[root].data = origin[index++];
	InOrder(node[root].right);
}

int num = 0;
void LevelOrder(int root) {
	queue<int> q;
	q.push(root);
	while (!q.empty()) {
		int top = q.front();
		printf("%d", node[top]);
		num++;
		if (num < n)
			printf(" ");
		q.pop();
		if (node[top].left != -1)
			q.push(node[top].left);
		if (node[top].right != -1)
			q.push(node[top].right);
	}
}

int main() {
	scanf("%d", &n);
	int lchild, rchild;
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &lchild, &rchild);
		node[i].left = lchild;
		node[i].right = rchild;
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", &origin[i]);
	}
	sort(origin, origin + n);
	InOrder(0);
	LevelOrder(0);

	return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值