Coding Test/LeetCode
[LeetCode] 606. Construct String from Binary Tree c++
owls
2022. 12. 24. 14:58
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- 문제
Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.
Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.
Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
P (LC)(RC) 형태로 부모 노드와 연결된 자식노드들을 괄호로 표시하는 문제이다.
RC가 없다면 해당 괄호는 생략해도 된다 : P(LC)
LC가 없다면 해당 괄호는 생략하면 안된다. P()(RC)
- 문제 해결
/**
* 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 {
private:
void tree2str(TreeNode* curr, string& str){
if(!curr){
return;
}
str += to_string(curr->val);
if(curr->left){
str += '(';
tree2str(curr->left, str);
str += ')';
}
if(curr->right){
if(!curr->left){
str += "()";
}
str += '(';
tree2str(curr->right, str);
str += ')';
}
return;
}
public:
string tree2str(TreeNode* root) {
string str("");
tree2str(root, str);
return str;
}
};
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