剑指offer--面试题28：对称的二叉树

#include <cstdio>#define nullptr NULLstruct BinaryTreeNode {    int                    m_nValue;     BinaryTreeNode*        m_pLeft;      BinaryTreeNode*        m_pRight; };BinaryTreeNode* CreateBinaryTreeNode(int value){    BinaryTreeNode* pNode = new BinaryTreeNode();    pNode->m_nValue = value;    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);    }}bool isSymmetrical(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);bool isSymmetrical(BinaryTreeNode* pRoot){    return isSymmetrical(pRoot, pRoot);}bool isSymmetrical(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2){    if(pRoot1 == nullptr && pRoot2 == nullptr)        return true;    if(pRoot1 == nullptr || pRoot2 == nullptr)        return false;    if(pRoot1->m_nValue != pRoot2->m_nValue)        return false;    return isSymmetrical(pRoot1->m_pLeft, pRoot2->m_pRight)        && isSymmetrical(pRoot1->m_pRight, pRoot2->m_pLeft);}void main(){    BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);    BinaryTreeNode* pNode61 = CreateBinaryTreeNode(6);    BinaryTreeNode* pNode9 = CreateBinaryTreeNode(6);    BinaryTreeNode* pNode51 = CreateBinaryTreeNode(5);    BinaryTreeNode* pNode71 = CreateBinaryTreeNode(7);    BinaryTreeNode* pNode72 = CreateBinaryTreeNode(7);    BinaryTreeNode* pNode52 = CreateBinaryTreeNode(5);    ConnectTreeNodes(pNode8, pNode61, pNode9);    ConnectTreeNodes(pNode61, pNode51, pNode71);    ConnectTreeNodes(pNode9, pNode72, pNode52);if(isSymmetrical( pNode8))printf("是对称的二叉树\n");    DestroyTree(pNode8);}