Sub-structure and sub-structure of the tree

Sub-structure and sub-structure of the tree

1. Question

Enter Two Binary Trees A and B to determine whether B is A sub-structure. The binary tree structure is as follows:

Struct BinaryTreeNode

{

Int m_nValue;

BinaryTreeNode * m_pLeft;

BinaryTreeNOde * m_pRight;

};

In example 1-1 (a), the red part and (B) have the same structure: the corresponding data fields are the same.

Figure 1-1 substructure

2. Analysis

Obviously, if the root node pRoot1 and pRoot2 of Tree B in tree A meet the following requirements, B is the sub-structure of:

(1) The values of pRoot1 and pRoot2 are the same;

(2) The left subtree of pRoot1 is exactly the same as that of pRoot2;

(3) The right subtree of pRoot1 is exactly the same as that of pRoot2.

3. Implementation Method

Method 1:

The result is obtained through recursive traversal tree. The time complexity O (m * n), where m and n are the number of nodes in A and B respectively.

Method 2:

(1) traverse tree A and B in order to get two traversal sequences: Apre and Bpre.

(2) The next traversal tree A and B get two traversal sequences: Apast and Bpast.

(3) Use the string comparison algorithm KMP for detection. If Apre contains Bpre and Apast contains Bpast, B is the sub-structure of A, and vice versa.

The time complexity O (m + n) of this method, where m and n are the number of nodes in A and B respectively.

Note: because a traversal order cannot uniquely determine a binary tree, two traversal sequences are required.

4. Code

In the Code, if the two operators are null, B is not the sub-structure of.

boolHasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2){    bool result = false;     if(pRoot1 != NULL && pRoot2 !=NULL)    {        if(pRoot1->m_nValue ==pRoot2->m_nValue)            result = DoesTree1HaveTree2(pRoot1,pRoot2);        if(!result)            result =HasSubtree(pRoot1->m_pLeft, pRoot2);        if(!result)            result = HasSubtree(pRoot1->m_pRight,pRoot2);    }     return result;} boolDoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2){    if(pRoot2 == NULL)        return true;     if(pRoot1 == NULL)        return false;     if(pRoot1->m_nValue != pRoot2->m_nValue)        return false;     return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) && DoesTree1HaveTree2(pRoot1->m_pRight,pRoot2->m_pRight);}