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C++算法之 求二叉树中叶子节点的个数 与 判断两棵二叉树是否结构相同

//叶子节点的个数
/*
(1)如果二叉树为空,返回0
(2)如果二叉树不为空且左右子树为空,返回1
(3)如果二叉树不为空,且左右子树不同时为空,返回左子树中叶子节点个数加上右子树中叶子节点个数
*/

int GetLeafNodeNum(BTree* root){	if(root == NULL)		return 0;	if(root->m_pLeft == NULL && root->m_pRight == NULL)		return 1;	int LeafNumOfLeft = GetLeafNodeNum(root->m_pLeft);	int LeafNumOfRight = GetLeafNodeNum(root->m_pRight);	int ret = LeafNumOfLeft + LeafNumOfRight;	return ret;}


 

/*
判断量个二叉树的结构是否相同
1:如果两个二叉树都为空,那么返回true
2:如果一个二叉树为空,另外一个不为空,那么返回false
3:如果两个二叉树都不为空,那么如果它们的左子树和右子树的结果相同,返回true否则返回false

*/

bool isEqual(BTree*  root1, BTree* root2){	if(root1 == NULL && root2 == NULL)		return true;	else if ((root1 == NULL && root2!= NULL)|| (root1 != NULL && root2 == NULL))		return false;	bool equalLeft = isEqual(root1->m_pLeft,root2->m_pLeft);	bool equalRight = isEqual(root1->m_pRight,root2->m_pRight);	return (equalLeft && equalRight);}


完整测试代码:

// BTNumOfKLevel.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>using namespace std;class BTree{public:	int     m_nValue;	BTree*  m_pLeft;	BTree*  m_pRight;	BTree(int m):m_nValue(m)	{		m_pLeft = m_pRight = NULL;	}};//二叉树的插入实现void Insert(int value, BTree* &root){	if (root == NULL)	{		root = new BTree(value);	}	else if(value < root->m_nValue)		Insert(value,root->m_pLeft);	else if(value > root->m_nValue)		Insert(value,root->m_pRight);	else		;}//叶子节点的个数/*(1)如果二叉树为空,返回0(2)如果二叉树不为空且左右子树为空,返回1(3)如果二叉树不为空,且左右子树不同时为空,返回左子树中叶子节点个数加上右子树中叶子节点个数*/int GetLeafNodeNum(BTree* root){	if(root == NULL)		return 0;	if(root->m_pLeft == NULL && root->m_pRight == NULL)		return 1;	int LeafNumOfLeft = GetLeafNodeNum(root->m_pLeft);	int LeafNumOfRight = GetLeafNodeNum(root->m_pRight);	int ret = LeafNumOfLeft + LeafNumOfRight;	return ret;}/*判断量个二叉树的结构是否相同1:如果两个二叉树都为空,那么返回true2:如果一个二叉树为空,另外一个不为空,那么返回false3:如果两个二叉树都不为空,那么如果它们的左子树和右子树的结果相同,返回true否则返回false*/bool isEqual(BTree*  root1, BTree* root2){	if(root1 == NULL && root2 == NULL)		return true;	else if ((root1 == NULL && root2!= NULL)|| (root1 != NULL && root2 == NULL))		return false;	bool equalLeft = isEqual(root1->m_pLeft,root2->m_pLeft);	bool equalRight = isEqual(root1->m_pRight,root2->m_pRight);	return (equalLeft && equalRight);}int _tmain(int argc, _TCHAR* argv[]){	BTree* m_pRoot = new BTree(4);	Insert(3,m_pRoot);	Insert(6,m_pRoot);	Insert(1,m_pRoot);	Insert(2,m_pRoot);	Insert(5,m_pRoot);	Insert(8,m_pRoot);	Insert(7,m_pRoot);	Insert(10,m_pRoot);	BTree* m_pRoot2 = new BTree(4);	Insert(3,m_pRoot2);	Insert(6,m_pRoot2);	Insert(1,m_pRoot2);	Insert(2,m_pRoot2);	Insert(5,m_pRoot2);	Insert(8,m_pRoot2);	Insert(7,m_pRoot2);	Insert(10,m_pRoot2);	int count = GetLeafNodeNum(m_pRoot);	cout<<"叶子节点的个数为:"<<count<<endl;	cout<<"两个树的结构是否相同:"<<isEqual(m_pRoot,m_pRoot2);	getchar();	return 0;}


 

 

 

 

C++算法之 求二叉树中叶子节点的个数 与 判断两棵二叉树是否结构相同