PAT A1099. Build A Binary Search Tree (30) [⼆叉查找树BST]

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本文介绍了一种基于二叉搜索树（BST）的层序遍历算法实现。通过定义二叉树结构和使用递归填充节点值，文章详细解释了如何根据输入的树结构和一组唯一整数键来构建并输出该二叉搜索树的层序遍历序列。此过程涉及中序遍历以保持BST特性，并最终按层级顺序输出节点。

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.
在这里插入图片描述
题⽬⼤意：给出⼀棵⼆叉搜索树（给出每个结点的左右孩⼦），且已知根结点为0，求并且给出应该插
⼊这个⼆叉搜索树的数值，求这棵⼆叉树的层序遍历
分析： 1. ⽤结构体data， lef， right， index， level表示这棵树的结构， a数组存树的信息， b数组存这棵
树节点的所有data，根据输⼊可知树a[i]的lef和right
2. 因为是⼆叉搜索树，所以中序遍历这棵树得到的结点顺序应该是给出的数值序列从⼩到⼤的排
列顺序，所以把数值序列排序后，可以在中序遍历的时候直接赋值当前tree[root].data～
3. 然后根据节点的层数和下标，就可以排序输出层序

#include <iostream>
#include <algorithm>
using namespace std;
int n, cnt, b[100];
struct node {
	int data, l, r, index, lebel;
}a[110];
bool cmp(node x, node y) {
	if (x.lebel != y.lebel) return x.lebel < y.lebel;
	return x.index < y.index;
}
void dfs(int root, int index, int lebel) {
	if (a[root].l == -1 && a[root].r == -1) {
		a[root] = {b[cnt++], a[root].l, a[root].r, index, lebel};
	} else {
		if (a[root].l != -1) dfs(a[root].l, index * 2 + 1, lebel + 1);
		a[root] = {b[cnt++], a[root].l, a[root].r, index, lebel};
		if (a[root].r != -1) dfs(a[root].r, index * 2 + 2, lebel + 1);
	}
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i].l >> a[i].r;
	for (int i = 0; i < n; i++)
		cin >> b[i];
	sort(b, b + n);
	dfs(0, 0, 0);
	sort(a, a + n, cmp);
	for (int i = 0; i < n; i++) {
		if (i != 0) cout << " ";
		cout << a[i].data;
	}
	return 0;
}