A1110. Complete Binary Tree

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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<string.h>
 #include<queue>
 using namespace std;
 typedef struct{
     int lchild, rchild;
 }node;
 node tree[];
 int str2num(char ss[]){
     int len = strlen(ss);
     if(ss[] == '-')
         return -;
     int P = , ans = ;
     for(int i = len - ; i >= ; i--){
         ans += (ss[i] - '') * P;
         P *= ;
     }
     return ans;
 }
 int N, lastNode, root, notRoot[] = {}, cnt = , tag = ;
 void levelOrder(int root){
     queue<int> Q;
     Q.push(root);
     while(Q.empty() == false){
         int temp = Q.front();
         lastNode = temp;
         Q.pop();
         cnt++;
         if(cnt < N /  && (tree[temp].lchild == - || tree[temp].rchild == -)){
             tag = ;
             return;
         }else if(cnt == N /  && (tree[temp].lchild == - && tree[temp].rchild == - || tree[temp].lchild == - && tree[temp].rchild != -)){
             tag = ;
             return;
         }else if(cnt > N /  && (tree[temp].lchild != - || tree[temp].rchild != -)){
             tag = ;
             return;
         }
         if(tree[temp].lchild != -)
             Q.push(tree[temp].lchild);
         if(tree[temp].rchild != -)
             Q.push(tree[temp].rchild);
     }
 }
 int main(){
     scanf("%d", &N);
     char str[];
     for(int i = ; i < N; i++){
         scanf("%s", str);
         tree[i].lchild = str2num(str);
         scanf("%s", str);
         tree[i].rchild = str2num(str);
         if(tree[i].lchild != -)
             notRoot[tree[i].lchild] = ;
         if(tree[i].rchild != -)
             notRoot[tree[i].rchild] = ;
     }
     for(int i = ; i < N; i++){
         if(notRoot[i] == ){
             root = i;
             break;
         }
     }
     levelOrder(root);
     if(tag == )
         printf("NO %d", root);
     else printf("YES %d", lastNode);
     cin >> N;
     return ;
 }

总结：

1、题目要求判断二叉树是否是完全二叉树。我的办法是二叉树的层序遍历，访问一个节点就cnt++。访问前 N/2 - 1个节点时要求必须都有左右孩子。第N/2个节点要求必须是左右都非空或左非空右为空。N/2之后的节点要求左右子树都必须空。

2、网上看到还有更简单的方法，就是在层序遍历的时候把为-1的空节点也都加入队列。当访问时，遇到-1节点则查看cnt，如果cnt<N 则说明不是完全二叉树。另外注意，最后一个非空节点不一定是N-1。因为虽然是完全二叉树，但它的层次遍历节点序号不是按照0、1、2、3的顺序。