【C++】满二叉树问题

/*
给出一棵满二叉树的先序遍历，有两种节点：字母节点（A-Z，无重复）和空节点（#）。要求这个树的中序遍历。输出中序遍历时不需要输出#。
满二叉树的层数n满足1<=n<=5。Sample Input:
ABC#D#ESample Output:
CBADE
*/
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
using namespace std;
const int M=1024;
char data[M];
int layer;struct node
{char data;struct node *l;struct node *r;
};void build(node * & t,int l)
{if(l>layer)return ;int i=0;while(data[i]=='\0')i++;char tc=data[i];//cout<<i<<endl<<tc<<endl;
    data[i]='\0';if(tc=='#'){t=NULL;return ;}t=new node();t->data=tc;build(t->l,l+1);build(t->r,l+1);
}void display(node *t)
{if(t==NULL)return ;display(t->l);printf("%c",t->data);display(t->r);
}int main()
{/*printf("%d\n",(int)(log(8)/log(2)));printf("%f\n",(log(7)/log(2)));*/memset(data,'\0',sizeof(data));while(scanf("%s",&data)){//printf("%s\n",data);int n=strlen(data)+1;/*if(log(n)/log(2)>(int)(log(n)/log(2)))layer=(int)(log(n)/log(2))+1;else*/layer=(int)(log(n)/log(2));//printf("%d\n",layer);node *tree;build(tree,1);display(tree);}return 0;
}

 

 

tz@HZAU

2019/3/13