二叉树的叶子节点数和树的高度

问题：这两种操作都用到了递归，别人说递归好理解，我真的不觉得递归好理解，只是递归的代码看起来简单。

下面代码是求高度的另一种方法。递归计算高度是从底向上计算的，因此叶子节点高度为0；

方法真的很巧妙。。。

比如getTreeHigh(a->lchild)返回的就是a的左子树的高度，getTreeHigh(a->lchild)返回的就是a的右子树的高度。

int getTreeHigh(BinTree btree)
{
    int depth;
        if(btree==NULL)
        depth=0;
    else
    {
        depthleft=getTreeHigh(btree->lchild);
        depthright=getTreeHigh(btree->rchild);
        depth=1+max(depthleft,depthright);
    }
    return depth;
}

代码：

#include <iostream>
#include <cstdlib>
#include <stack>
using namespace std;
static int num=0;
typedef struct btree
{
    char data;
    struct btree *lchild;
    struct btree *rchild;
}*BinTree;
 
void CreateBinTree(BinTree &btree)
{
    char c;
    cin>>c;
    if(c=='#')
        btree=NULL;
    else
    {
        btree=(BinTree)malloc(sizeof(struct btree));
        btree->data=c;
        CreateBinTree(btree->lchild);
        CreateBinTree(btree->rchild);
    }
}
void showBTree(BinTree btree)
{
    if(btree)
    {
        cout<<btree->data<<" ";
        showBTree(btree->lchild);
        showBTree(btree->rchild);
    }
}
 
int getTreeHigh(BinTree btree)
{
       if(btree==NULL)
        return 0;
    if(btree->lchild&&btree->rchild)
        return max(getTreeHigh(btree->lchild)+1,getTreeHigh(btree->rchild)+1);
    if(btree->lchild)
        return getTreeHigh(btree->lchild)+1;
    if(btree->rchild)
        return getTreeHigh(btree->rchild)+1;
    return 1;
 
}
 
int getNumOfleaves(BinTree btree)
{
 
    if(btree)
    {
    if(btree->lchild==NULL&&btree->rchild==NULL)
        num++;
    else
    {
        getNumOfleaves(btree->lchild);
        getNumOfleaves(btree->rchild);
    }
    }
    return num;
}
int main()
{
    BinTree bt;
    int count;
    int h=0;
    cout<<"create bintree:"<<endl;
    CreateBinTree(bt);
    cout<<"output the bintree:"<<endl;
    showBTree(bt);
    cout<<endl;
    count=getNumOfleaves(bt);
    cout<<"叶子节点数为: "<<count<<endl;
    h=getTreeHigh(bt);
    cout<<"二叉树的高度为: "<<h<<endl;
    return 0;
}