pat甲级1099 build a Binary Search Tree

1099 Build A Binary Search Tree (30 分)

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

/**
题意 已经将搜索二叉树的下标索引位置给出，给出具体的值，需要将值填入搜索二叉树中， 并层序输出二叉搜索树
 
 
这道题的思路就是
1.建立二叉搜索树
2.二叉搜索树建立完后，将二叉搜索树的值插入对应索引处（因为二叉搜索树的中序遍历就是从小到大排序，因此只需
将具体的值从小到大排序后 并将排序后的值经中序遍历输入到二叉搜索树中 (并根据二叉树的性质 root(root) = 0 左index = 2 * index + 1 右index = 2 * 2 + 2）
将结点送入到对应的向量容器中（向量容器根据index从小到排序），（但此题还需要按先将高度从大到小排序）  ） 按顺序输出向量容器中的值 就是层序遍历
 
 
**/

方法一：

             用 下标索引输出层序遍历：

             具体代码：

            

#include<iostream>
#include<algorithm>
#include<vector>
/**
题意 已经将搜索二叉树的下标索引位置给出，给出具体的值，需要将值填入搜索二叉树中， 并层序输出二叉搜索树
 
 
这道题的思路就是
1.建立二叉搜索树
2.二叉搜索树建立完后，将二叉搜索树的值插入对于索引处（因为二叉搜索树的中序遍历就是从小到大排序，因此只需
将具体的值从小到大排序后 并将排序后的值经中序遍历输入到二叉搜索树中 (并根据二叉树的性质 root(root) = 0 左index = 2 * index + 1 右index = 2 * 2 + 2） 
将结点送入到对应的向量容器中（向量容器根据index从小到排序），（但此题还需要按先将高度从大到小排序）  ） 按顺序输出向量容器中的值 就是层序遍历 
 
 
**/

using namespace std;
struct Node{
	int index, l, r, data, h;
}node[100];
int n, cnt = 0;
vector<int> in;
vector<Node> level;
void inOrder(int root, int index, int h){ //中序遍历 将已有的中序遍历结点依此放入对应搜索BST的索引中 
	if(root == -1) return;
	inOrder(node[root].l, index * 2 + 1, h + 1);
	level.push_back({index, node[root].l, node[root].r, in[cnt++], h});
	inOrder(node[root].r, index * 2 + 2, h + 1); 
}
bool cmp(Node a, Node b){
	if(a.h != b.h)
		return a.h < b.h;
	return a.index < b.index;
}
int main(){
	int l, r, data;
	scanf("%d", &n);
	in.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d%d", &l, &r);
		node[i].l = l;
		node[i].r = r;
	}
	for(int i = 0; i < n; i++){
		scanf("%d", &data);
		in[i] = data;
	}
	sort(in.begin(), in.end());
	inOrder(0, 0, 0);
	sort(level.begin(), level.end(), cmp);
	for(int i = 0; i < level.size(); i++){
		if(i != 0)
			printf(" ");
		printf("%d", level[i].data);
	}	
	return 0;
} 

方法2：

            运用队列遍历层序遍历

           具体代码：

#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
/**
printf("%s%d", k++ == 0 ? "" : " ", n.data); 三目运算符中 第一个输出的是字符,需要在printf表示出来 
**/
using namespace std;
struct Node{
	int data, l, r;
}node[100];
queue<Node> q;
int n, cnt = 0;
vector<int> in;
void BSTorder(int root){
	if(node[root].l != -1) BSTorder(node[root].l);
	node[root].data = in[cnt++];
	if(node[root].r != -1) BSTorder(node[root].r);
}
int main(){
	int l, r, k = 0;
	scanf("%d", &n);
	in.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d%d", &l, &r);
		node[i].l = l;
		node[i].r = r;
	}
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]);
	}
	sort(in.begin(), in.end());
	BSTorder(0);
	q.push(node[0]);         //层序遍历 
	while(!q.empty()){
		Node n = q.front(); //取出队首结点 
		q.pop();
		if(n.l != -1)
			q.push(node[n.l]);
		if(n.r != -1)
			q.push(node[n.r]);
		printf("%s%d", k++ == 0 ? "" : " ", n.data);
	}
	
		
	return 0;
}