判断一棵二叉树是不是平衡二叉树

 1 bool IsBalanced(BinaryTreeNode* root)
 2 {
 3     if (root == NULL)
 4     {
 5         return true ;
 6     }
 7     int left = TreeDepth(root->m_pLeft);//该函数实现见我上一篇博客“数的深度”
 8     int right = TreeDepth(root->m_pRight);
 9     if ((left - right) > 1 || (left - right) < -1)
10     {
11         return false ;
12     }
13     else
14         return IsBalanced(root->m_pLeft) && IsBalanced(root->m_pRight);
15 }

 1 bool IsBalanced2(BinaryTreeNode* root , int& Depth)
 2 {
 3     if (root == NULL)
 4     {
 5         Depth = 0 ;
 6         return true ;
 7     }
 8     int left , right ;
 9     if (IsBalanced2(root->m_pLeft, left) && IsBalanced2(root->m_pRight, right))
10     {
11         int diff = left - right ;
12         if (diff <= 1 && diff >= -1)
13         {
14             Depth = 1 + (left > right ? left : right );
15             return true ;
16         }
17     }
18 
19     return false ;
20 
21 }
22 bool IsBalanced2(BinaryTreeNode* root)
23 {
24     int Depth = 0 ;
25     return IsBalanced2(root , Depth);
26 }