1099. Build A Binary Search Tree (30)

​ 时间限制

​ 100 ms

​ 内存限制

​ 65536 kB

​ 代码长度限制

​ 16000 B

​ 判题程序

​ Standard

​ 作者

​ CHEN, Yue

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.

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
 

此题为给你一颗构造好的BST，让你放置数，然后输出层次遍历

代码如下

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
struct Node{
	int key;
	int leftchild;
	int rightchild;
}node[101];
vector <int> v;
int index1=0;
void  bfs(int root,int index){
	if(node[root].leftchild!=-1){
		bfs(node[root].leftchild,2*index+1);
	}
	node[root].key=v[index1++];
	if(node[root].rightchild!=-1){
		bfs(node[root].rightchild,2*index+2);
	}
}
int main(){	
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;++i){
		scanf("%d%d",&node[i].leftchild,&node[i].rightchild);
	}
	v.resize(n);
	for(int i=0;i<n;++i){
		scanf("%d",&v[i]);
	}
	sort(v.begin(),v.end());
	bfs(0,0);
	queue<int>q;
	q.push(0);
	printf("%d",node[0].key);
	while(!q.empty()){
		int root=q.front();
		q.pop();
		if(root!=0) printf(" %d",node[root].key);
		if(node[root].leftchild!=-1) q.push(node[root].leftchild);
		if(node[root].rightchild!=-1) q.push(node[root].rightchild);
	}
}