判断一棵二叉树是否是平衡二叉树/求一颗二叉树的镜像

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);}