题解 | #单链表的排序#

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
        // 当树为空时，返回TRUE
        if(pRoot == nullptr) 
            return true;
        int leftHeight = treeHeight(pRoot->left);
        int rightHeight = treeHeight(pRoot->right);
        if(abs(leftHeight - rightHeight) > 1)  // 判断左右子树的差值是否大于一，大于一返回FALSE
            return false;
        // 最后当左子树和右子树都符合平衡二叉树时，返回TRUE
        return IsBalanced_Solution(pRoot->left) && IsBalanced_Solution(pRoot->right);



    }
    int treeHeight(TreeNode* pRoot) { // 计算左右子树的层数
        if(pRoot==nullptr)
            return 1;
        int leftHeight = treeHeight(pRoot->left);
        int rightHeight = treeHeight(pRoot->right);
        return max(leftHeight, rightHeight) + 1;
    }
};