PAT-A1110 Complete Binary Tree 题目内容及题解

Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤20) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1

题目大意

题目给定一棵树，要求判断它是否是一个完整的二叉树。如是则返回最后一个节点的编号；如果不是就返回根节点的编号。

解题思路

1. 读入树并将出现过的节点进行记录；
2. 找到未出现过的节点即为根；
3. 通过前序遍历观察节点序号是否始终在给定值以内，如是则判断其为完全二叉树，否则不是；
4. 按照题目要求输出并返回零值。

代码

#include<stdio.h>
struct Node{
    int lchild,rchild;
}node[30];

int flag=0;
int N,root,Max,isroot[30];

void Init(){
    int i,j,num;
    char a[3],b[3];
    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%s%s",a,b);
        if(a[0]=='-'){
            node[i].lchild=-1;
        }else{
            j=0,num=0;
            while(a[j]!=0){
                num=num*10+a[j]-'0';
                j++;
            }
            node[i].lchild=num;
            isroot[num]=1;
        }
        if(b[0]=='-'){
            node[i].rchild=-1;
        }else{
            j=0,num=0;
            while(b[j]!=0){
                num=num*10+b[j]-'0';
                j++;
            }
            node[i].rchild=num;
            isroot[num]=1;
        }
    }
    for(i=0;i<N;i++){
        if(isroot[i]==0){
            root=i;
            break;
        }
    }
}

void PreOrder(int now,int seq){
    if(flag||now==-1){
        return;
    }
    if(seq>N){
        flag=1;
        return;
    }else if(seq==N){
        Max=now;
    }
    PreOrder(node[now].lchild,seq*2);
    PreOrder(node[now].rchild,seq*2+1);
}

int main(){
    Init();
    PreOrder(root,1);
    if(flag){
        printf("NO %d\n",root);
    }else{
        printf("YES %d\n",Max);
    }
}

运行结果