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;}