判断二叉树是否平衡

/** 
 * 题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树
 * 中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树
 */

int is_balanced(struct node * root_node, int * depth)
{
    if(root_node == NULL)
    {
        *depth = 0;
        return 1;
    }

    int left, right;

    if(is_balanced(root_node->left_node, &left) && \
        is_balanced(root_node->right_node, &right))
    {
        int diff = left - right;

        if(diff <= 1 && diff >= -1)
        {
            *depth = (left > right ? left : right) + 1;
            return 1;
        }
    }

    return 0;
}