题解 | 判断是不是平衡二叉树

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <ios>
class Solution {
private:
    int height(TreeNode* root){
        if(root == nullptr) return 0;
        return max(height(root->left),height(root->right))+1;
    }
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pRoot TreeNode类 
     * @return bool布尔型
     */
    bool IsBalanced_Solution(TreeNode* pRoot) {
        if(pRoot == nullptr) return true;
        int leftHeight = height(pRoot->left);
        int rightHeight = height(pRoot->right);
        if(abs(leftHeight - rightHeight) > 1) return false;
        return IsBalanced_Solution(pRoot->left) &&IsBalanced_Solution(pRoot->right);
        
    }
};