《每日一题》110. 平衡二叉树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int gethigh(TreeNode* node){
        if (node == NULL) return 0;
        int lhigh = 1 + gethigh(node->left);
        int rhigh = 1 + gethigh(node->right);
        int high = max(lhigh, rhigh);
        return high;
    }
    bool isBalanced(TreeNode* root) {
        if (root == NULL) return true;
        int lhigh = gethigh(root->left);
        int rhigh = gethigh(root->right);
        if (abs(lhigh - rhigh) > 1 ) return false;
        return isBalanced(root->left) && isBalanced(root->right);//每一个左右子树都要平衡
    }
};