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📚专栏地址:PAT题解集合
📝原题地址:题目详情 - 1099 Build A Binary Search Tree (pintia.cn)
🔑中文翻译:构建二叉搜索树
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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
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
Sample Output:
58 25 82 11 38 67 45 73 42
- 1
二叉搜索树 (BST) 递归定义为具有以下属性的二叉树:
给定二叉树的具体结构以及一系列不同的整数,只有一种方法可以将这些数填充到树中,以使结果树满足 BST 的定义。
请你输出结果树的层序遍历。
给定一个 N N N ,所有结点范围在 0 ∼ N − 1 0 \sim N-1 0∼N−1 之间,接下来 N N N 行会输入 0 ∼ N − 1 0 \sim N-1 0∼N−1 的左右孩子的下标,其中 − 1 -1 −1 表示空结点。
最后一行输出 N N N 个结点的值。
#include
using namespace std;
const int N = 110;
int n;
int l[N], r[N]; //用哈希表来存储每个结点左孩子的下标
int a[N], w[N];
int q[N];
//模拟中序遍历
void dfs(int u, int& k)
{
if (u == -1) return;
dfs(l[u], k);
w[u] = a[k++];
dfs(r[u], k);
}
//输出层次遍历的结果
void bfs()
{
int hh = 0, tt = 0;
q[0] = 0;
while (hh <= tt)
{
int t = q[hh++];
if (l[t] != -1) q[++tt] = l[t];
if (r[t] != -1) q[++tt] = r[t];
}
cout << w[q[0]];
for (int i = 1; i < n; i++) cout << " " << w[q[i]];
cout << endl;
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n); //获得中序遍历结果
int k = 0;
dfs(0, k);
bfs();
return 0;
}