面试题28：对称的二叉树

题目：请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。

#include<iostream>
using namespace std;
struct BinaryTreeNode {
	int m_nValue;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;
};
//思路：对称树的前序遍历和对称前序遍历相同
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);
}
//测试补充代码
BinaryTreeNode* CreateBinaryTreeNode(double dbValue)
{
	BinaryTreeNode* pNode = new BinaryTreeNode();
	pNode->m_nValue = 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);
	}
}
int main() {
	BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
	BinaryTreeNode* pNode61 = CreateBinaryTreeNode(6);
	BinaryTreeNode* pNode62 = CreateBinaryTreeNode(6);
	BinaryTreeNode* pNode51 = CreateBinaryTreeNode(5);
	BinaryTreeNode* pNode71 = CreateBinaryTreeNode(7);
	BinaryTreeNode* pNode72 = CreateBinaryTreeNode(7);
	BinaryTreeNode* pNode52 = CreateBinaryTreeNode(5);

	ConnectTreeNodes(pNode8, pNode61, pNode62);
	ConnectTreeNodes(pNode61, pNode51, pNode71);
	ConnectTreeNodes(pNode62, pNode72, pNode52);

	bool res = IsSymmetrical(pNode8, pNode8);
	DestroyTree(pNode8);
	return 0;
}