pat 1110 Complete Binary Tree (25分) 判断一棵二插树是否是完全二叉树

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 (≤) 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:

?

1 2 3 4 5 6 7 8 9 10

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

 

Sample Output 1:

?

1

YES 8

 

Sample Input 2:

?

1 2 3 4 5 6 7 8 9

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

Sample Output 2:

?

1

NO 1

 

解题思路：

这道题开始提交，有三个点通过不了，都显示段错误，想了半天，不应该段错误啊，原来是最开始输入，我用的是char输入，如果节点编号 >= 10那么char读不出来。应该用strng 输入，然后判断是否是" - "，是的话存为-1，否则正常存。

 

还有就是，判断完全二叉树，用层序遍历，如果前n个遍历序列没有出现-1，说明是完全二叉树，如果前n个序列出现了-1，则表示不是完全二叉树。代码如下

 

?

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#include<cstdio> #include<cstdlib> #include<cstring> #include<vector> #include<queue> #include<algorithm>   using namespace std; int tree[30];   int find(int x){  //寻找节点的根节点     if(tree[x] == x){         return x;     }     return find(tree[x]); }   int main(){     freopen("C:\\Users\\zzloyxt\\Desktop\\1.txt","r",stdin);        int n;     scanf("%d",&n);     for(int i=0;i<n;i++){  //存放每个下标节点的父亲节点。         tree[i] = i;     }     int left[n];     int right[n];     char a[3],b[3];     for(int i=0;i<n;i++){         scanf("%s %s",a,b);         if(strcmp(a,"-") ==0){             left[i] = -1;         }else{             int a_int = 0;             sscanf(a,"%d",&a_int);             left[i] = a_int;             tree[a_int] = i;         }         if(strcmp(b,"-") ==0){             right[i] = -1;         }else{             int b_int = 0;             sscanf(b,"%d",&b_int);  //将字符串转成整数。             right[i] = b_int;             tree[b_int] = i;         }     }     int root = find(0);     int shu[100] = {0};  //存放层序遍历序列     int index = 0;     queue<int> Q;     Q.push(root);     while(!Q.empty()){         int x = Q.front();         Q.pop();         shu[index++] = x;         if(x != -1){             Q.push(left[x]);             Q.push(right[x]);         }     }     int flag = 0;     for(int i=0;i<n;i++){         if(shu[i] < 0){             flag = 1;             break;         }     }     if(flag ==0){         printf("YES %d\n",shu[n-1]);     }else{         printf("NO %d\n",root);     }           return 0; }