判断一个二叉树是否平衡

判断二叉树是否平衡的方法是求左右儿子的深度相差是否大于1,大于则这个二叉树不平衡.

class Solution {
    public:
        int height(TreeNode *root) {//求节点的高度
            if(root == NULL)return 0;
            return max(height(root->left), height(root->right)) + 1;
        }
        bool isBalanced(TreeNode* root) {
            if(root == NULL)return true;
            return isBalanced(root->left) && isBalanced(root->right) && abs(height(root->left) - height(root->right)) <= 1;
        }
};