判断一棵树是否为平衡二叉树

Consider a height-balancing scheme where following conditions should be checked to determine if a binary tree is balanced.
An empty tree is height-balanced. A non-empty binary tree T is balanced if:

1. Left subtree of T is balanced
2. Right subtree of T is balanced
3. The difference between heights of left subtree and right subtree is not more than 1.

/* Returns true if binary tree with root as root is height-balanced */
bool isBalanced(struct node *root)
{
   int lh; /* for height of left subtree */
   int rh; /* for height of right subtree */ 
 
   /* If tree is empty then return true */
   if(root == NULL)
    return 1;
 
   /* Get the height of left and right sub trees */
   lh = height(root->left);
   rh = height(root->right);
 
   if( abs(lh-rh) <= 1 &&
       isBalanced(root->left) &&
       isBalanced(root->right))
     return 1;
 
   /* If we reach here then tree is not height-balanced */
   return 0;
}
  
/*  The function Compute the "height" of a tree. Height is the
    number of nodes along the longest path from the root node
    down to the farthest leaf node.*/
int height(struct node* node)
{
   /* base case tree is empty */
   if(node == NULL)
       return 0;
 
   /* If tree is not empty then height = 1 + max of left
      height and right heights */
   return 1 + max(height(node->left), height(node->right));
}