LeetCode in Kotlin
606. Construct String from Binary Tree
Easy
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
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
fun tree2str(t: TreeNode?): String {
if (t == null) {
return ""
}
val sb = StringBuilder()
preorder(t, sb)
return sb.toString()
}
private fun preorder(root: TreeNode?, sb: StringBuilder) {
if (root == null) {
return
}
sb.append(root.`val`)
if (root.left != null) {
sb.append("(")
preorder(root.left, sb)
sb.append(")")
}
if (root.right != null) {
if (root.left == null) {
sb.append("()")
}
sb.append("(")
preorder(root.right, sb)
sb.append(")")
}
}
}