https://leetcode.com/problems/construct-string-from-binary-tree/description/
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
- code
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private StringBuilder res = new StringBuilder();
public void dfs(TreeNode root){
if (root == null) return;
res.append(Integer.toString(root.val));
if (root.left == null && root.right == null) return;
res.append('(');
dfs(root.left);
res.append(')');
if (root.right != null){
res.append('(');
dfs(root.right);
res.append(')');
}
}
public String tree2str(TreeNode root) {
dfs(root);
return res.toString();
}
}