PAT甲级1099

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本文介绍了一道PAT甲级1099题目，要求根据给定的二叉搜索树结构和一串整数，通过中序遍历填充二叉树，并输出层序遍历结果。文章提供了完整的C++代码实现。

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PAT甲级1099

原题：
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

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

题目大意：给出总结点数n，然后给出第i个节点的左右孩子，根从0开始，若是-1则到底了。求出层序序列
由于题目说的是BST，所以给出序列从小到大排序即为中序序列，可以通过中序建树，然后再进行层序遍历即可，因为根始终从0开始所以不用额外求根

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

vector<int> in,level;
struct node{
	int data;	//节点存放的数据
	int left,right;	//左孩子右孩子
};
node nd[101];
int cnt=0;
void inOrder(int root){
	if(root==-1) return ;
	inOrder(nd[root].left);
	nd[root].data=in[cnt++];
	inOrder(nd[root].right);
}
void levelOrder(int root){
	if(root==-1) return ;
	queue<int> q;
	q.push(root);
	while(!q.empty()){
		int tmp=q.front();
		q.pop();
		level.push_back(tmp);
		if(nd[tmp].left!=-1) q.push(nd[tmp].left);
		if(nd[tmp].right!=-1) q.push(nd[tmp].right);
	}
}
int main(){
	int n,tmpleft,tmpright,tmp;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d%d",&tmpleft,&tmpright);
		nd[i].left=tmpleft;
		nd[i].right=tmpright;
	}
	for(int i=0;i<n;i++){
		scanf("%d",&tmp);
		in.push_back(tmp);
	}
	sort(in.begin(),in.end());
	inOrder(0);
	levelOrder(0);
	for(int i=0;i<level.size();i++){
		if(i!=0) printf(" ");
		printf("%d",nd[level[i]].data);
	}
	system("pause");
	return 0;
}