# 606. 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

## Solutions

• /**
* 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 {
public String tree2str(TreeNode root) {
if (root == null) {
return "";
}
if (root.left == null && root.right == null) {
return root.val + "";
}
if (root.right == null) {
return root.val + "(" + tree2str(root.left) + ")";
}
return root.val + "(" + tree2str(root.left) + ")(" + tree2str(root.right) + ")";
}
}

• /**
* 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) {
if (!root) return "";
return to_string(root->val) + "(" + tree2str(root->left) + ")(" + tree2str(root->right) + ")";
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def tree2str(self, root: Optional[TreeNode]) -> str:
def dfs(root):
if root is None:
return ''
if root.left is None and root.right is None:
return str(root.val)
if root.right is None:
return f'{root.val}({dfs(root.left)})'
return f'{root.val}({dfs(root.left)})({dfs(root.right)})'

return dfs(root)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func tree2str(root *TreeNode) string {
if root == nil {
return ""
}
if root.Left == nil && root.Right == nil {
return strconv.Itoa(root.Val)
}
if root.Right == nil {
return strconv.Itoa(root.Val) + "(" + tree2str(root.Left) + ")"
}
return strconv.Itoa(root.Val) + "(" + tree2str(root.Left) + ")(" + tree2str(root.Right) + ")"
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function tree2str(root: TreeNode | null): string {
if (root == null) {
return '';
}
if (root.left == null && root.right == null) {
return ${root.val}; } return ${root.val}(${root.left ? tree2str(root.left) : ''})${
root.right ? (\${tree2str(root.right)}) : ''
};
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut String) {
if let Some(node) = root {
let node = node.borrow();
res.push_str(node.val.to_string().as_str());

if node.left.is_none() && node.right.is_none() {
return;
}
res.push('(');
if node.left.is_some() {
Self::dfs(&node.left, res);
}
res.push(')');
if node.right.is_some() {
res.push('(');
Self::dfs(&node.right, res);
res.push(')');
}
}
}

pub fn tree2str(root: Option<Rc<RefCell<TreeNode>>>) -> String {
let mut res = String::new();
Self::dfs(&root, &mut res);
res
}
}