pat1099. Build A Binary Search Tree (30)

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.
• 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
  

---

提交代码

 

平衡二叉树。每次找到中间节点。

#include<cstdio>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<queue>
#include<map>
using namespace std;
struct node{
    int l,r,v;
};
node tree[105];
int line[105];
int calnum(int num){
    if(num==-1){
        return 0;
    }
    return calnum(tree[num].l)+calnum(tree[num].r)+1;
    }
void BuildAVL(int mid,int *line){
    if(mid==-1){
        return ;
    }
    int count=calnum(tree[mid].l);//统计左子树

    //cout<<count<<endl;

    tree[mid].v=line[count];
    BuildAVL(tree[mid].l,line);
    BuildAVL(tree[mid].r,line+count+1);
}
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    int n;
    scanf("%d",&n);
    int i;
    for(i=0;i<n;i++){
        scanf("%d %d",&tree[i].l,&tree[i].r);
    }
    for(i=0;i<n;i++){
        scanf("%d",&line[i]);
    }
    sort(line,line+n);

    /*for(i=0;i<n;i++){
        cout<<i<<" "<<line[i]<<endl;
    }*/

    BuildAVL(0,line);

    //cout<<":"<<"  "<<tree[0].v<<endl;

    queue<int> q;
    int cur;
    q.push(0);
    printf("%d",tree[0].v);
    while(!q.empty()){
        cur=q.front();
        q.pop();
        //cout<<"cur:  "<<cur<<endl;

        if(tree[cur].l!=-1){
            q.push(tree[cur].l);
            printf(" %d",tree[tree[cur].l].v);
        }
        if(tree[cur].r!=-1){
            q.push(tree[cur].r);
            printf(" %d",tree[tree[cur].r].v);
        }
    }
    printf("\n");
    return 0;
}