LeetCode in Kotlin
101. Symmetric Tree
Easy
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
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 isSymmetric(root: TreeNode?): Boolean {
return if (root == null) {
true
} else {
helper(root.left, root.right)
}
}
private fun helper(leftNode: TreeNode?, rightNode: TreeNode?): Boolean {
if (leftNode == null || rightNode == null) {
return leftNode == null && rightNode == null
}
return if (leftNode.`val` != rightNode.`val`) {
false
} else {
helper(leftNode.left, rightNode.right) && helper(leftNode.right, rightNode.left)
}
}
}