C语言每日一题（56）平衡二叉树

输入：root = [3,9,20,null,null,15,7]
输出：true

输入：root = [1,2,2,3,3,null,null,4,4]
输出：false

输入：root = []
输出：true

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 int BinaryTreeHight(struct TreeNode* root)//求二叉树高度
{
	if (root == NULL)
	{
		return 0;
	}
	return fmax(BinaryTreeHight(root->left), BinaryTreeHight(root->right)) + 1;
	
}
bool isBalanced(struct TreeNode* root) {
    if(root==NULL)
    {
        return true;
    }
    int left=BinaryTreeHight(root->left);//保存左子树高度
    int right=BinaryTreeHight(root->right);//保存右子树高度
    int max=left;//假设法
    int min=right;
    if(left<right)
    {
        min=left;
        max=right;
    }
    if((max-min)>1)
    {
        return false;
    }
    return isBalanced(root->left)&&isBalanced(root->right);
    
}