1110 Complete Binary Tree (25分)测试点2,3,4

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.

分析:用结构体存储每个结点，如果左孩子或右孩子为空用-1代替，之后根据完全二叉树的特点将结点全部存入数组中。

测试点2，3，4：因为结点可能是两位数，我之前用char类型去获取左右孩子结点，但是结点可能是两位数，所以需要用string

AC代码:

#include<iostream>
#include<algorithm>
using namespace std;
struct node{int left;int right;
}tree[20];
int res[20];
int n;
void dfs(int root,int index){if(index>n){return;}res[index]=root;if(tree[root].left!=-1){dfs(tree[root].left,index*2+1);}if(tree[root].right!=-1){dfs(tree[root].right,index*2+2);}return;
}
int main(){cin>>n;bool flag[n];fill(flag,flag+n,false);fill(res,res+20,-1);for(int i=0;i<n;++i){string a,b;//一定要注意节点可能是两位数 cin>>a>>b;if(a[0]=='-'){tree[i].left=-1;}else{int t=stoi(a);tree[i].left=t;flag[t]=true;}if(b[0]=='-'){tree[i].right=-1;}else{int t=stoi(b);tree[i].right=t;flag[t]=true;}}int root=0;for(int i=0;i<n;++i){//找根 if(flag[i]==false){root=i;break;}}dfs(root,0);bool tt=false;for(int i=0;i<n;++i){if(res[i]==-1){tt=true;break;}}if(tt){cout<<"NO"<<" "<<res[0];}else{cout<<"YES"<<" "<<res[n-1];}return 0;
}

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1110 Complete Binary Tree (25分)测试点2,3,4

测试点2，3，4：因为结点可能是两位数，我之前用char类型去获取左右孩子结点，但是结点可能是两位数，所以需要用string

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