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.
Example 1:
Input: root = [1,2,3,4] Output: "1(2(4))(3)" Explanation: Originally, it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)"
Example 2:
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.
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -1000 <= Node.val <= 1000
Companies: Amazon, Adobe, Facebook
Related Topics:
String, Tree, Depth-First Search, Binary Tree
Similar Questions:
Plain DFS. Notice when you should add the left/right child part.
// OJ: https://leetcode.com/problems/construct-string-from-binary-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
string tree2str(TreeNode* root) {
if (!root) return "";
auto val = to_string(root->val);
if (!root->left && !root->right) return val;
return val + "(" + tree2str(root->left) + ")" + (root->right ? "(" + tree2str(root->right) + ")" : "");
}
};