1：判断是否为平衡二叉树：

//方法1：
int TreeDepth(BTree* pRoot)
{
	if (pRoot == NULL)
		return 0;
	int nLeftDepth = TreeDepth(pRoot->m_pLeft);
	int nRightDepth = TreeDepth(pRoot->m_pRight);

	return (nLeftDepth > nRightDepth)? (nLeftDepth+1):(nRightDepth+1);
}

bool IsBalanced(BTree* pRoot)
{
	if (pRoot == NULL)
		return true;
	int nLeftDepth = TreeDepth(pRoot->m_pLeft);
	int nRightDepth = TreeDepth(pRoot->m_pRight);
	int diff = nRightDepth - nLeftDepth;
	if (diff > 1 || diff < -1)
		return false;

	return IsBalanced(pRoot->m_pLeft)&&IsBalanced(pRoot->m_pRight);
	
}
/*
上面的方法：在求该节点的左右子树的深度的时候遍历一遍树，再次判断子树的平衡性的时候又要遍历
一遍树结构，造成遍历多次！
*/
bool IsBalanced3(BTree* pRoot, int& depth)
{
	if(pRoot== NULL)
	{
		depth = 0;
		return true;
	}

	int nLeftDepth;
	bool bLeft= IsBalanced3(pRoot->m_pLeft, nLeftDepth);
	int nRightDepth;
	bool bRight = IsBalanced3(pRoot->m_pRight, nRightDepth);
	
	if (bLeft && bRight && abs(nLeftDepth - nRightDepth) <=1)
	{
		depth = 1+(nLeftDepth > nRightDepth ? nLeftDepth : nRightDepth);
		return true;	
	}
	else
	{
		return false;
	}
}

bool IsBalanced3(BTree* pRoot)
{
	int depth = 0;
	return IsBalanced3(pRoot, depth);
}

2：求二叉树的镜像

/*
求二叉树的镜像：

方法1： 前序遍历每个节点，如果遍历到的节点有子节点，就交换它的两个子节点。（先交换左子树和右子树，再对左子树和右子树进行镜像操作）
方法2： 如果二叉树不为空，求左子树和右子树的镜像，然后再交换左子树和右子树

*/

void Mirror(BTree* &pRoot)
{
	if(pRoot == NULL)
		return;
	if(pRoot->m_pLeft ==NULL && pRoot->m_pRight == NULL)
		return;

	BTree* pTemp = pRoot->m_pLeft;
	pRoot->m_pLeft = pRoot->m_pRight;
	pRoot->m_pRight = pTemp;

	if(pRoot->m_pLeft)
		Mirror(pRoot->m_pLeft);
	if(pRoot->m_pRight)
		Mirror(pRoot->m_pRight);
}

BTree* Mirror2(BTree* pRoot)
{
	if(pRoot == NULL)
		return NULL;

	BTree*  pLeft = Mirror2(pRoot->m_pLeft);
	BTree*  pRight = Mirror2(pRoot->m_pRight);


	pRoot->m_pLeft = pRight;
	pRoot->m_pRight = pLeft;

	return pRoot;
}

完整测试代码：