判断一棵二叉树是否是平衡二叉树

判断每个节点的做右子树高度差，递归法

求一颗二叉树的镜像

交换左右孩子节点

template<class T>
struct TreeNode
{
    TreeNode* _Lchild;
    TreeNode* _Rchild;
    T data;
};
size_t Hight(TreeNode* pRoot)
{
    if (pRoot==NULL)
    {
        return 0;
    }
    size_t lhight=Hight(pRoot->_Lchild)+1;
    size_t rhight=Hight(pRoot->_Rchild)+1;
    return lhight>rhight ? lhight:rhight;
}
bool IsBanlence(TreeNode* pRoot)
{
    if (pRoot==NULL)
    {
        return true;
    }
    int Lhight=Hight(pRoot->_Lchild);
    int Rhight=Hight(pRoot->_Rchild);
    int hight=Lhight-Rhight;
    if (hight>=-1 && hight<=1)
    {
        return IsBanlence(pRoot->_Lchild)&&
               IsBanlence(pRoot->_Rchild);
    }
    else
        return false;
}
void Mirror(TreeNode* pRoot)//镜像
{
    if (pRoot==NULL)
    {
        return ;
    }
    if (pRoot->_Lchild==NULL && pRoot->_Rchild==NULL)
    {
        return ;
    }
    swap(pRoot->_Rchild,pRoot->_Lchild);
    Mirror(pRoot->_Rchild);
    Mirror(pRoot->_Lchild);
}