## Algorithm

Problem Name: 606. Construct String from Binary Tree

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 Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
string tree2str(TreeNode* t) {
if(!t) return "";
string l = tree2str(t->left);
string r = tree2str(t->right);
return to_string(t->val) + "(" + l + ")" + "(" + r + ")";
}
};
``````
Copy The Code &

Input

cmd
root = [1,2,3,4]

Output

cmd
"1(2(4))(3)"

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public String tree2str(TreeNode root) {
StringBuilder sb = new StringBuilder();
helper(root, sb);
return sb.toString();
}

private void helper(TreeNode root, StringBuilder sb) {
if (root == null) {
return;
}
sb.append(root.val);
if (root.left == null && root.right == null) {
return;
}
sb.append("(");
helper(root.left, sb);
sb.append(")");
if (root.right != null) {
sb.append("(");
helper(root.right, sb);
sb.append(")");
}
}
}
``````
Copy The Code &

Input

cmd
root = [1,2,3,4]

Output

cmd
"1(2(4))(3)"

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const tree2str = function(t) {
if (!t) return ''
const left = tree2str(t.left)
const right = tree2str(t.right)
if (right) return `\${t.val}(\${left})(\${right})`
else if (left) return `\${t.val}(\${left})`
else return `\${t.val}`
};
``````
Copy The Code &

Input

cmd
root = [1,2,3,null,4]

Output

cmd
"1(2()(4))(3)"

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def tree2str(self, t):
"""
:type t: TreeNode
:rtype: str
"""
if not t: return ""
parent="%s" %t.val
left, right= "", ""
if t.left or t.right: left= "(%s)" % self.tree2str(t.left)
if t.right: right= "(%s)" % self.tree2str(t.right)
return parent+left+right
``````
Copy The Code &

Input

cmd
root = [1,2,3,null,4]

Output

cmd
"1(2()(4))(3)"

### #5 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0606_ConstructStringFromBinaryTree
{
public string Tree2str(TreeNode t)
{
if (t == null) return string.Empty;
if (t.left == null && t.right == null) return \$"{t.val}";
if (t.right == null) return \$"{t.val}({Tree2str(t.left)})";
return \$"{t.val}({Tree2str(t.left)})({Tree2str(t.right)})";
}
}
}
``````
Copy The Code &

Input

cmd
root = [1,2,3,4]

Output

cmd
"1(2(4))(3)"