PAT (Advanced Level) 1099. Build A Binary Search Tree (30)

Posted Fighting Heart

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了PAT (Advanced Level) 1099. Build A Binary Search Tree (30)相关的知识,希望对你有一定的参考价值。

预处理每个节点左子树有多少个点。

然后确定值得时候递归下去就可以了。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

struct Node
{
    int val;
    int left;
    int right;
    int numL,numR;
}s[200];
int n,a[200];
vector<int>ans;

void dfs(int x)
{
    if(s[x].left!=-1)
    {
        dfs(s[x].left);
        s[x].numL=s[s[x].left].numL+s[s[x].left].numR+1;
    }

    if(s[x].right!=-1)
    {
        dfs(s[x].right);
        s[x].numR=s[s[x].right].numL+s[s[x].right].numR+1;
    }
}

void Find(int x,int L,int R)
{
    s[x].val=a[L+s[x].numL];
    if(s[x].left!=-1) Find(s[x].left,L,L+s[x].numL-1);
    if(s[x].right!=-1) Find(s[x].right,L+s[x].numL+1,R);
}

void bfs()
{
    queue<int>q;
    q.push(0);
    while(!q.empty())
    {
        int h=q.front(); q.pop();
        ans.push_back(s[h].val);
        if(s[h].left!=-1) q.push(s[h].left);
        if(s[h].right!=-1) q.push(s[h].right);
    }

}
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++) {
        scanf("%d%d",&s[i].left,&s[i].right);
        s[i].numL=s[i].numR=0;
    }
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+1+n);
    dfs(0);
    Find(0,1,n);
    bfs();
    for(int i=0;i<ans.size();i++)
    {
        printf("%d",ans[i]);
        if(i<ans.size()-1) printf(" ");
        else printf("\n");
    }
    return 0;
}

 

以上是关于PAT (Advanced Level) 1099. Build A Binary Search Tree (30)的主要内容,如果未能解决你的问题,请参考以下文章

PAT (Advanced Level) 1099. Build A Binary Search Tree (30)

PAT Advanced 1099 Build A Binary Search Tree (30分)

PAT (Advanced Level) 1025. PAT Ranking (25)

PAT Advanced 1099 Build A Binary Search Tree (30) [?叉查找树BST]

PAT Advanced Level 1044

PAT Advanced Level 1043