LeetCode - Binary Tree Inorder Traversal

Given the root of a binary tree, return the inorder traversal of its nodes' values.

 

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

 

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

풀이(Kotlin)

 - 재귀를 사용하여 중위순회(left -> root -> right)를 구하였다.

/**
 * 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 inorderTraversal(root: TreeNode?): List<Int> {
        val list = ArrayList<Int>()
        inorder(root, list)
        return list
    }
    fun inorder(node: TreeNode?, list: MutableList<Int>) {
        if (node == null) return
        inorder(node.left, list)
        list.add(node.`val`)
        inorder(node.right, list)
    }
}