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;
}

运行结果：