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 <utility>
class Solution {
  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
     */
    pair<bool, int> IsBalancedFun(TreeNode* pRoot) {
        if (pRoot == nullptr) {
            return make_pair(true, 0);
        }
        pair<bool, int> left, right;
        left = IsBalancedFun(pRoot->left);
        right = IsBalancedFun(pRoot->right);
        return make_pair(left.first && right.first &&
                         (left.second - right.second <= 1 &&
                          left.second - right.second >= -1),
                         left.second > right.second ? left.second + 1 : right.second + 1);
    }
    bool IsBalanced_Solution(TreeNode* pRoot) {
        // write code here
        return IsBalancedFun(pRoot).first;
    }
};