解题思路:
改动的二叉树中序遍历,具体表现为加括号时同时检查左右子树是否为空。
代码:
/**
* 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:
string tree2str(TreeNode* root) {
string str = to_string(root -> val);
if(root -> left != NULL || root -> right != NULL) {
str += (root -> left == NULL ? "()" : '(' + tree2str(root -> left) + ')') + (root -> right == NULL ? "" : '(' + tree2str(root -> right) + ')');
}
return str;
}
};