LeetCode-in-All
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
/*
* Definition for a binary tree node.
* class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
* var value: Int = _value
* var left: TreeNode = _left
* var right: TreeNode = _right
* }
*/
object Solution {
def isSymmetric(root: TreeNode): Boolean = {
if (root == null) {
true
} else {
helper(root.left, root.right)
}
}
private def helper(leftNode: TreeNode, rightNode: TreeNode): Boolean = {
if (leftNode == null || rightNode == null) {
leftNode == null && rightNode == null
} else if (leftNode.value != rightNode.value) {
false
} else {
helper(leftNode.left, rightNode.right) && helper(leftNode.right, rightNode.left)
}
}
}