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);
if(!t->left && !t->right) return to_string(t->val);
if(!t->right) return to_string(t->val) + "(" + l + ")";
return to_string(t->val) + "(" + l + ")" + "(" + r + ")";
}
};
Input
Output
#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(")");
}
}
}
Input
Output
#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}`
};
Input
Output
#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
Input
Output
#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)})";
}
}
}
Input
Output