判断一颗二叉树是否是平衡二叉树

方法一，参考 http://blog.csdn.net/forestlight/article/details/6575507

 template<typename T>
 int DepthTree(BSTreeNode<T> *pbs)
 {
     if (pbs==NULL)
         return 0;
     else
     {
         int leftLength=DepthTree(pbs->left);
         int rigthLength=DepthTree(pbs->right);
         return 1+(leftLength>rigthLength ? leftLength:rigthLength);
     }
 }

 template<typename T>
 bool isBalanceTree(BSTreeNode<T> *pbs)
 {
     if (pbs==NULL)
     {
         return true;
     }

     int depthLeft=DepthTree(pbs->left);
     int depthRight=DepthTree(pbs->right);

     if (abs(depthLeft-depthRight)>1)
          return false;
     else
         return isBalanceTree(pbs->left) && isBalanceTree(pbs->right);
 }

方法一用的是线序遍历的思想， 缺点是，重复计算了多次子树的深度

方法二用的是后序遍历的思想，免去了对子树的重复计算

template<typename T>  
bool isBalanceTree(BSTreeNode<T> *pbs, int &curLen)  
{  
    if (pbs==NULL)  
    {  
        return true;  
    }  
      
    int leftLen;
    int rightLen;
    bool isLeftBalance;
    bool isRightBalance;
	
    leftLen = rightLen = 0;
    isLeftBalance = isRightBalance = false;
	
    isLeftBalance  = isBalanceTree(pbs->left, leftLen);  
    isRightBalance = isBalanceTree(pbs->right, rightLen);  
      
    curLen =  leftLen >= rightLen ?  leftLen+1 : rightLen+1;
	  
    if (isLeftBalance == true && isRightBalance == true)
    {
        if (abs(depthLeft-depthRight) <= 1) 
        {
            return true;
        }
    }	
    return false;
}