剑指offer面试题26 ：树的子结构

提交网址: 
http://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170或 http://www.lintcode.com/zh-cn/problem/subtree/ (难度: Easy)

题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。(ps：我们约定空树不是任意一个树的子结构)

思路：

要判断二叉树 A 中是否存在和树 B 结构一样的子结构，可以分成两步： 
1.在树 A 中找到和 B 的根结点的值一样的结点R; 
2.再判断树 A 中以 R 为根结点的子树是不是包含和树 B 一样的结构。
AC代码：

/*
1.输入两颗二叉树A,B 判断B是不是A的子结构
我们约定空树不是任意一个树的子结构
*/
#include<iostream>
#include<vector>
using namespace std;
struct TreeNode
{
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x):
val(x),left(NULL),right(NULL){}
};
class Solution
{
public:
bool HasSubtree(TreeNode* pRoot1,TreeNode *pRoot2)
{
bool res=false;
if(pRoot1!=NULL && pRoot2!=NULL)
{
if(pRoot1->val==pRoot2->val) res=Tree1HaveTree2(pRoot1,pRoot2);
if(res==false) res=HasSubtree(pRoot1->left,pRoot2);
if(res==false) res=HasSubtree(pRoot1->right,pRoot2);
}
return res;
}
bool Tree1HaveTree2(TreeNode *pRoot1,TreeNode *pRoot2)
{
if(pRoot2==NULL) return true; //进行到这一步的时候 说明在这一步的节点之前 都符合

if(pRoot1==NULL) return false;//前两个递归出口条件的顺序有影响 到这一步的时候 说明pRoot2!=nullptr 而pRoot1=nullptr 所以return false
if(pRoot1->val!=pRoot2->val) return false;
return Tree1HaveTree2(pRoot1->left,pRoot2->left) && Tree1HaveTree2(pRoot1->right,pRoot2->right);
}
};
int main()
{
Solution sol;
TreeNode *pRoot1=new TreeNode(1);
pRoot1->left=new TreeNode(2);
pRoot1->right=new TreeNode(3);
pRoot1->left->left=new TreeNode(4);
pRoot1->left->right=new TreeNode(5);

/*
A树
1
/ \
2 3
/ \
4 5
*/

TreeNode *pRoot2=new TreeNode(2);
pRoot2->left=new TreeNode(4);
pRoot2->right=new TreeNode(5);
/*
B树
2
/ \
4 5
*/

if( (sol.HasSubtree(pRoot1,pRoot2) )==true)
{
cout<<"true";
}
else if( (sol.HasSubtree(pRoot1,pRoot2) )==false)
{
cout<<"false";
}

}
看看自己测试的方法简直太单调太low

这是剑指offer作者完整源码：还需要多看看/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt) *******************************************************************/

//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================

// 面试题26：树的子结构
// 题目：输入两棵二叉树A和B，判断B是不是A的子结构。

#include <cstdio>

struct BinaryTreeNode
{
double m_dbValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
};

bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);
bool Equal(double num1, double num2);

bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
bool result = false;

if(pRoot1 != nullptr && pRoot2 != nullptr)
{

cbae
if(Equal(pRoot1->m_dbValue, pRoot2->m_dbValue))
result = DoesTree1HaveTree2(pRoot1, pRoot2);
if(!result)
result = HasSubtree(pRoot1->m_pLeft, pRoot2);
if(!result)
result = HasSubtree(pRoot1->m_pRight, pRoot2);
}

return result;
}

bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
if(pRoot2 == nullptr)
return true;

if(pRoot1 == nullptr)
return false;

if(!Equal(pRoot1->m_dbValue, pRoot2->m_dbValue))
return false;

return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
}

bool Equal(double num1, double num2)
{
if((num1 - num2 > -0.0000001) && (num1 - num2 < 0.0000001))
return true;
else
return false;
}

// ====================辅助测试代码====================
BinaryTreeNode* CreateBinaryTreeNode(double dbValue)
{
BinaryTreeNode* pNode = new BinaryTreeNode();
pNode->m_dbValue = dbValue;
pNode->m_pLeft = nullptr;
pNode->m_pRight = nullptr;

return pNode;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
if(pParent != nullptr)
{
pParent->m_pLeft = pLeft;
pParent->m_pRight = pRight;
}
}

void DestroyTree(BinaryTreeNode* pRoot)
{
if(pRoot != nullptr)
{
BinaryTreeNode* pLeft = pRoot->m_pLeft;
BinaryTreeNode* pRight = pRoot->m_pRight;

delete pRoot;
pRoot = nullptr;

DestroyTree(pLeft);
DestroyTree(pRight);
}
}

// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2, bool expected)
{
if(HasSubtree(pRoot1, pRoot2) == expected)
printf("%s passed.\n", testName);
else
printf("%s failed.\n", testName);
}

// 树中结点含有分叉，树B是树A的子结构
// 8 8
// / \ / \
// 8 7 9 2
// / \
// 9 2
// / \
// 4 7
void Test1()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA6 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNodeA7 = CreateBinaryTreeNode(7);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);

Test("Test1", pNodeA1, pNodeB1, true);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树中结点含有分叉，树B不是树A的子结构
// 8 8
// / \ / \
// 8 7 9 2
// / \
// 9 3
// / \
// 4 7
void Test2()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeA6 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNodeA7 = CreateBinaryTreeNode(7);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);

Test("Test2", pNodeA1, pNodeB1, false);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树中结点只有左子结点，树B是树A的子结构
// 8 8
// / /
// 8 9
// / /
// 9 2
// /
// 2
// /
// 5
void Test3()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, nullptr);
ConnectTreeNodes(pNodeA2, pNodeA3, nullptr);
ConnectTreeNodes(pNodeA3, pNodeA4, nullptr);
ConnectTreeNodes(pNodeA4, pNodeA5, nullptr);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, pNodeB2, nullptr);
ConnectTreeNodes(pNodeB2, pNodeB3, nullptr);

Test("Test3", pNodeA1, pNodeB1, true);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树中结点只有左子结点，树B不是树A的子结构
// 8 8
// / /
// 8 9
// / /
// 9 3
// /
// 2
// /
// 5
void Test4()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, nullptr);
ConnectTreeNodes(pNodeA2, pNodeA3, nullptr);
ConnectTreeNodes(pNodeA3, pNodeA4, nullptr);
ConnectTreeNodes(pNodeA4, pNodeA5, nullptr);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);

ConnectTreeNodes(pNodeB1, pNodeB2, nullptr);
ConnectTreeNodes(pNodeB2, pNodeB3, nullptr);

Test("Test4", pNodeA1, pNodeB1, false);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树中结点只有右子结点，树B是树A的子结构
// 8 8
// \ \
// 8 9
// \ \
// 9 2
// \
// 2
// \
// 5
void Test5()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, nullptr, pNodeA3);
ConnectTreeNodes(pNodeA3, nullptr, pNodeA4);
ConnectTreeNodes(pNodeA4, nullptr, pNodeA5);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, nullptr, pNodeB3);

Test("Test5", pNodeA1, pNodeB1, true);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树A中结点只有右子结点，树B不是树A的子结构
// 8 8
// \ \
// 8 9
// \ / \
// 9 3 2
// \
// 2
// \
// 5
void Test6()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, nullptr, pNodeA3);
ConnectTreeNodes(pNodeA3, nullptr, pNodeA4);
ConnectTreeNodes(pNodeA4, nullptr, pNodeA5);

BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeB4 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, pNodeB3, pNodeB4);

Test("Test6", pNodeA1, pNodeB1, false);

DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}

// 树A为空树
void Test7()
{
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeB4 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, pNodeB3, pNodeB4);

Test("Test7", nullptr, pNodeB1, false);

DestroyTree(pNodeB1);
}

// 树B为空树
void Test8()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);

ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, pNodeA3, pNodeA4);

Test("Test8", pNodeA1, nullptr, false);

DestroyTree(pNodeA1);
}

// 树A和树B都为空
void Test9()
{
Test("Test9", nullptr, nullptr, false);
}

int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
Test9();

return 0;
}