Build A Binary Search Tree

题目描述

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.

 

 

输入描述:

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.

 

输出描述:

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.

 

输入例子:

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 

 

输出例子:

58 25 82 11 38 67 45 73 42

考察二叉树的中序与层次遍历，注意传参数时传值与传引用的区别：

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=110;
struct node{
	int data,left=-1,right=-1;
}nd[N];
int num[N];
int n;
//根据排序好的数组，中序遍历将数据插入 
void midsort(int root,int &index){
	if(root==-1) return;//遇到子节点，直接返回 
	midsort(nd[root].left,index);//递归遍历左子树
	nd[root].data=num[index++];
	//index++;
	midsort(nd[root].right,index);//递归遍历右子树 
} 
//输出层序遍历的结果
void output(int root){
	queue<int>q;
	q.push(root);
	int flag=1;
	while(!q.empty()){
		if(!flag) cout<<" ";
		int t=q.front();
		//cout<<t<<endl;
		q.pop();
		cout<<nd[t].data;
		flag=0;
		if(nd[t].left!=-1){
			q.push(nd[t].left);
		}
		if(nd[t].right!=-1){
			q.push(nd[t].right);
		}
	}
}
int main(){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d%d",&nd[i].left,&nd[i].right);
	}
	for(int i=0;i<n;i++)
		scanf("%d",&num[i]);
	sort(num,num+n);
	int index;
	midsort(0,index);
	output(0);
	
	return 0;
}