C++练习 | 递归创建二叉树并求叶子结点的数值和

#include <iostream>
using namespace std;

struct Tree
{
    int data;
    Tree *lchild;
    Tree *rchild;
}tree;

Tree *Create(int a1[],int b1[],int n)
{
    int k;
    if(n<=0)
        return NULL;
    int root=a1[0];
    Tree *bt=(Tree *)malloc(sizeof(Tree));
    bt->data=root;
    for(k=0;k<n;k++)
    {
        if(b1[k]==root)
            break;//分割左右子树
    }
    bt->lchild=Create(a1+1,b1,k);
    bt->rchild=Create(a1+k+1,b1+k+1,n-k-1);
    return bt;
}

int Tsum(Tree *r)
{
    if(r->lchild==NULL&&r->rchild==NULL)
            return r->data;
    if(r->lchild==NULL&&r->rchild!=NULL)
        return Tsum(r->rchild);
    if(r->lchild!=NULL&&r->rchild==NULL)
        return Tsum(r->lchild);
    if(r->lchild!=NULL&&r->rchild!=NULL)
        return Tsum(r->lchild)+Tsum(r->rchild);
    return r->data;
}

int main()
{
    Tree *rt;
    int sum1;
    int a[107],b[107],N;//N为总节点数，a为先序序列，b为中序序列
    cin>>N;
    for(int i=0;i<N;i++)
        cin>>a[i];
    for(int i=0;i<N;i++)
        cin>>b[i];
    rt=Create(a,b,N);//存树
    sum1=Tsum(rt);
    cout<<sum1<<endl;
    return 0;
}