第26题:LeetCode572:Subtree of Another Tree另一个树的子树

题目描述

给定两个非空二叉树 s 和 t,检验 s 中是否包含和 t 具有相同结构和节点值的子树。s 的一个子树包括 s 的一个节点和这个节点的所有子孙。s 也可以看做它自身的一棵子树。

示例 1:
给定的树 s:

     3
    / \
   4   5
  / \
 1   2

给定的树 t:

   4 
  / \
 1   2

返回 true,因为 t 与 s 的一个子树拥有相同的结构和节点值。

示例 2:
给定的树 s:

     3
    / \
   4   5
  / \
 1   2
    /
   0

给定的树 t:

   4
  / \
 1   2

返回 false

 

测试用例

// 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);
}

考点

1.二叉树遍历

2.递归

3.鲁棒性:每次访问指针要检查是否为nullptr,防御性编程

4.用几个测试用例进行单元测试。

 

思路

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);
    }
}

第一遍

//对于float和double型的小数,无法直接比较大小,判断这两个数相等的做法是:两数之差的绝对值在一个很小的范围内,0.0000001
bool Equal(double num1, double num2)
{ 
	return ((num1 - num2) > -0.000000001 && (num1 - num2) < 0.00000001) ? true : false; 
}

//递归检查根节点下的子树是否结构一样
bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{

	//技巧2:递归结束条件:大树为空,说明已经遍历完了,说明存在;子树为空,就返回假

	//1.如果B子树为空,返回真,说明已经遍历完子树了
	if (!pRoot2)
		return true;

	//2.如果A子树为空,返回假,说明不匹配 
	if (!pRoot1)
		return false;


	//3.如果两个节点不相同,返回假,说明不匹配 
	if (!Equal(pRoot1->m_dbValue, pRoot2->m_dbValue))
		return false;

	//4.返回递归调用,各自的左节点和各自的右节点递归
	return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) && DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
	 
}

//是否为子树
bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
	//1.如果有空树,返回假
	if (!pRoot1 || !pRoot2)
		return false;

	//2.默认返回结果为假
	bool result = false;
	
	//3.如果两个根节点相同就判断第二步
	if (Equal( pRoot1->m_dbValue,pRoot2->m_dbValue))//技巧1:小数用定义的函数判断,不用==判断相等。
		result = DoesTree1HaveTree2(pRoot1, pRoot2);
	if (!result)
		//如果不相同,就往左遍历,直到找到第一个相同的根节点
		result = HasSubtree(pRoot1->m_pLeft, pRoot2);
	if (!result)//如果不相同,就往右遍历,直到找到第一个相同的根节点
		result = HasSubtree(pRoot1->m_pRight, pRoot2);

	//4.返回结果
	return result;
}

第二遍(newcoder)

/*
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)
    {
        //1.如果有空树,直接返回
        if(!pRoot1||!pRoot2)
            return false;
        
        //2.bool 记录结果
        bool result=false;
        
        //3.如果根节点不相等,result=false
        if(!numCompare(pRoot1,pRoot2))
           result=false;
        else//3.1如果相等,就进行第二步
            result= HasSubtree2( pRoot1,  pRoot2);
        
        //4.如果result=false,遍历左子树的根节点是否与B树根节点相等
        if(!result)
            result=HasSubtree(pRoot1->left,pRoot2);
        //5.如果result=false,遍历右子树的根节点是否与B树根节点相等
        if(!result)
            result=HasSubtree(pRoot1->right,pRoot2);
        //6.返回result
        return result;
    }
    //递归
    bool HasSubtree2(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        //1.递归结束条件:B树为空,返回真
        if(!pRoot2)
            return true;
        //2.递归结束条件2:A树为空,返回假
        if(!pRoot1)
            return false;
        
        //3.如果这两个节点相等,分别继续判断左子树和右子树
        if(numCompare(pRoot1,pRoot2))
            return HasSubtree2(pRoot1->left,pRoot2->left)&&HasSubtree2(pRoot1->right,pRoot2->right);
        else 
            //如果不相等 返回假
            return false;
       
    }
    
    
    bool numCompare(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        return (pRoot1->val-pRoot2->val)<0.0000001&&(pRoot1->val-pRoot2->val)>-0.0000001?true:false;
    }
};

第三遍(leetcode572)

错误解答:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        //1.如果有一个为空,返回假
        if(!s||!t)
            return false;
        
        //2.记录结果
        bool result=false;
        
        //3.如果根节点相同
        if(compare(s,t))
            //进入第二步
            result=HasSubtree(s,t);
        
        //4.如果根节点不同,result=false,遍历左节点
        if(!result)
            result=isSubtree(s->left,t);
        
        //5.如果S的左树都没有相同的根节点,遍历右树
        if(!result)
            result=isSubtree(s->right,t);
        
        //6.返回result
        return result;
        
        
    }
    
    //第二步,递归函数,
    bool HasSubtree(TreeNode* s, TreeNode* t)
    {
        //1.如果子树遍历完,返回真
        if(!t)
            return true;
        
        //2.如果大树遍历完,返回假
        if(!s)
            return false;
        
        //3.比较根节点大小
        if(compare(s,t))
            //根节点相同,分别遍历左右子树,返回fun(left)&&fun(right)的结果
            return HasSubtree(s->left,t->left)&&HasSubtree(s->right,t->right);
        else
            //否则返回假
            return false;
    }
    
    
    //比较小数大小
    bool compare(TreeNode* s,TreeNode* t)
    {
        return (s->val-t->val)<0.0000001&&(s->val-t->val)>-0.0000001?true:false;     
}
};

//[3, 4, 5, 1, 2, null, null, 0]
//[4, 1, 2]
//              3
//            /    \
//          4         5
//         / \
//        1    2
//       /
//      0

[4,1,2]少了一个左孩子0 ,所以不是子树

Q:

为什么输入是
[1,2,3]
[1,2]
预期结果是:false呢? 

A:

因为【1,2】不是【1,2,3】的子树啊,【1,2,3】的子树是它自身和【2】,【3】


正解: 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
 
    bool isSameTree(TreeNode*s, TreeNode* t){
    if (!s) return !t;
    if (!t) return false;
    if (s->val != t->val) return false;
    return isSameTree(s->left, t->left) & isSameTree(s->right, t->right);
}

bool isSubtree(TreeNode* s, TreeNode* t) {
    if (!s) return !t;
    if (s->val == t->val && isSameTree(s, t))
        return true;
    return isSubtree(s->left, t) | isSubtree(s->right, t);
}
};

 最优解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode* s, TreeNode* t,bool check)
    {
        //1.都为空,返回真
        if(!s&&!t)
            return true;
        
        //2.只有一个为空,返回假
        if(!s || !t)
            return false;
        
        //3.如果第一步相同,第二步不同,返回假
         if(check&&s->val!=t->val)
             return false;
             
        bool ret=false;
        
        //4.如果根的值相同,&&分别递归遍历左右子树 
        if(s->val==t->val)            
           ret=isSameTree(s->left,t->left,true)&&isSameTree(s->right,t->right,true);
        
        //5.ret=ret或左子树或右子树,||
        ret=ret||isSameTree(s->left,t,false)||isSameTree(s->right,t,false);    
        
        //6.返回ret
        return ret;
        
    }
    
    bool isSubtree(TreeNode* s, TreeNode* t ) {
        return isSameTree(s,t,false);
    }
    
};

这道题还可以用序列化二叉树,然后用KMP方法做,树M,子树N,M>=N,时间复杂度O(M)。

后面再贴KMP的做法。

posted @ 2019-02-07 15:53 lightmare 阅读(...) 评论(...) 编辑 收藏