用DevC++，求二叉树叶结点个数，求不出来。不知道哪错了，求助大神!

Question

二叉树求叶结点求不出来

int LeafCount(BiTree T){
    if(!T)
    return 0;
    if(T!=0&&T->lchild==NULL&&T->rchild==NULL)
        return 1+LeafCount(T->lchild)+LeafCount(T->rchild);
    else 
    LeafCount(T->lchild)+LeafCount(T->rchild);
}

完整代码如下：

using namespace std;
typedef char TElemType;
#define MAXSIZE 50
#include <iostream>
typedef struct BiTNode{
    TElemType data;
    struct BiTNode *lchild,*rchild;
}BitNode,*BiTree;
void CreateBiTree(BiTree &T){
    TElemType ch;
    cin>>ch;
    if(ch=='#')  T=NULL;
    else{
        T=new BiTNode;
        T->data=ch;
        CreateBiTree(T->lchild);
        CreateBiTree(T->rchild);
    }
}
void PreOrderTraverse(BiTree T){
    if(T){
        cout<<T->data;
        PreOrderTraverse(T->lchild);
        PreOrderTraverse(T->rchild);
    }
}
void InOrderTraverse(BiTree T){
    if(T){
        InOrderTraverse(T->lchild);
        cout<<T->data;
        InOrderTraverse(T->rchild);
    }
}
void PostOrderTraverse(BiTree T){
    if(T){
        PostOrderTraverse(T->lchild);
        PostOrderTraverse(T->rchild);
        cout<<T->data;
    }
}
int Depth(BiTree T){
    int m,n;
    int depthval;
    if(!T)
        depthval=0;
    else{
        m=Depth(T->lchild);
        n=Depth(T->rchild);
        depthval=1+(m>n?m:n);
    }
    return depthval;
}
int NodeCount(BiTree T){
    if(!T)
    return 0;
    else
    return NodeCount(T->lchild)+NodeCount(T->rchild)+1;
}
int LeafCount(BiTree T){
if(!T)
return 0;
return LeafCount(T->lchild) + LeafCount(T->rchild) + 1;
}
/*int LeafCount(BiTree T){
    if(!T)
        return 0;
    if(T->lchild==NULL&&T->rchild==NULL)
        return 1+LeafCount(T->lchild)+LeafCount(T->rchild);
    else 
    LeafCount(T->lchild)+LeafCount(T->rchild);
}*/
int NodeNumber_1(BiTree T){
    if(!T)
    return 0;
    if((T->lchild==NULL&&T->rchild!=NULL)||(T->lchild!=NULL&&T->rchild==NULL))
        return 1+NodeNumber_1(T->lchild)+NodeNumber_1(T->rchild);
    else
        return NodeNumber_1(T->lchild)+NodeNumber_1(T->rchild);
}
void Route(BiTree T,char *path,int pathlen){  //输出二叉树中从每个叶子结点到根结点的路径
    if(T == NULL){
        return; 
    }
    if(T->lchild == NULL && T->rchild == NULL){
        cout<<endl<<T->data<<" ";
        for(int i = pathlen-1; i >= 0; i--){
            cout<<path[i]<<" ";
        }   
    }
    else{
        path[pathlen++] = T->data;
        Route(T->lchild,path,pathlen);
        Route(T->rchild,path,pathlen);
    }
}

int main()
{
    BiTree T;
    cout<<"请先序遍历输入(以#结束):";
    CreateBiTree(T);
    cout<<"中序遍历输出:";
    InOrderTraverse(T);
    cout<<endl<<"先序遍历输出:";
    PreOrderTraverse(T);
    cout<<endl<<"后序遍历输出:";
    PostOrderTraverse(T);
    cout<<endl<<"树的深度:"<<Depth(T);
    cout<<endl<<"结点的个数:"<<NodeCount(T);
    cout<<endl<<"叶结点的个数:"<<LeafCount(T);
    cout<<endl<<"度为1的结点个数:"<<NodeNumber_1(T);
    cout<<endl<<"二叉树中从每个叶子结点到根结点的所有路径："<<endl;
    char path[256];
    int pathlen=0;
    Route(T,path,pathlen);
    return 0;
}

Answer 1

int LeafCount(BiTree T){
if(!T)
return 0;
return LeafCount(T->lchild) + LeafCount(T->rchild) + 1;
}

问题解决请点下采纳，包括之前的 https://ask.csdn.net/questions/768205

我回答一个问题分文不取，花费很多时间，你点采纳仅仅举手之劳。