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二叉树,后续遍历
#include<iostream> using namespace std; string ans; int n; void build(int l,int r) { int mid=(r+l)/2; if(l!=r) { build(l,mid); build(mid+1,r); } int num0=0,num1=0; for(int i=l;i<=r;i++) { if(ans[i]=='0') num0++; else num1++; } if(r-l+1==num0) cout<<'B'; else if(r-l+1==num1) cout<<'I'; else cout<<'F'; } int main() { while(cin>>n>>ans) { n=1<<n; build(0,n-1); cout<<endl; } return 0; }
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原文地址:http://blog.csdn.net/zafkiel_nightmare/article/details/45371215