判断一颗树是否为平衡二叉树

struct BinaryTreeNode
{
    int m_Value;
    BinaryTreeNode* m_pLeft;
    BinaryTreeNode* m_pRight;
};
int height(BinaryTreeNode*node ,bool&balance)
{
if(node==NULL)
{
return 0;}
int R=node->right?height(node->right,balance)+1:0;
if(!balance)
return 0;
int l=node->left?height(node->left,balance)+1:0;
if(!balance)
return 0;
int dif=l-r;
if (dif<=1)
{
return (l>r?l:r)
}
}
bool  isBalancedTree(node *root)
{
bool balance=tree;
if(root)
  treehight(root,balance)
return balance;
}

 

 

 

 

 

 

//递归：

struct BinaryTreeNode
{
    int m_Value;
    BinaryTreeNode* m_pLeft;
    BinaryTreeNode* m_pRight;
};

int TreeDepth(BinaryTreeNode* 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);
}

 

判断该树是否为平衡二叉树

 

方法一：调用上述函数求每个节点的左右孩子深度

 

bool IsBalanced(BinaryTreeNode* 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);
}