力扣细节题：判断是否为平衡二叉树

经典题，需要记忆，且注意fabs和fmax函数的使用

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 int deep(struct TreeNode*root)
 {
     if(root==NULL)
     {
         return 0;
     }
     return fmax( deep(root->left),deep(root->right))+1;
 }
bool isBalanced(struct TreeNode* root) {
    if(root==NULL)
    {
        return true;
    }
    if(fabs(deep(root->left)-deep(root->right))>1)
    {
        return false;
    }
    if(isBalanced(root->left)&&isBalanced(root->right))
    {
        return true;
    }
    return false;
}