1099 Build A Binary Search Tree (30分)

题目地址

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

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=1010,inf=100000000;
struct node{
    int data,lchild,rchild;
}s[maxn];
int n,a[maxn],cnt=0;
void in(int idx){
    if(idx==-1) return;
    in(s[idx].lchild);
    s[idx].data=a[cnt++];
    in(s[idx].rchild);
}
void level(int idx){
    queue<node>q;
    q.push(s[idx]);int  i=0;
    while(!q.empty()){
        node now=q.front();
        if(now.lchild!=-1) q.push(s[now.lchild]);
        if(now.rchild!=-1) q.push(s[now.rchild]);
        if(i!=0) cout<<" ";
        cout<<now.data;
        i++;
        q.pop();
    }
}
int main(){
    cin>>n;
    for(int i=0;i<n;i++) cin>>s[i].lchild>>s[i].rchild;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    in(0);
    level(0);
}