LeetCode in Kotlin

623. Add One Row to Tree

Medium

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.

Note that the root node is at depth 1.

The adding rule is:

• Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur’s left subtree root and right subtree root.
• cur’s original left subtree should be the left subtree of the new left subtree root.
• cur’s original right subtree should be the right subtree of the new right subtree root.
• If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root’s left subtree.

Example 1:

Input: root = [4,2,6,3,1,5], val = 1, depth = 2

Output: [4,1,1,2,null,null,6,3,1,5]

Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3

Output: [4,2,null,1,1,3,null,null,1]

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• The depth of the tree is in the range [1, 104].
• -100 <= Node.val <= 100
• -105 <= val <= 105
• 1 <= depth <= the depth of tree + 1

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 addOneRow(root: TreeNode?, `val`: Int, depth: Int): TreeNode? {
        if (depth == 1) {
            val newRoot = TreeNode(`val`)
            newRoot.left = root
            return newRoot
        }
        dfs(root!!, depth - 2, `val`)
        return root
    }

    private fun dfs(node: TreeNode, depth: Int, `val`: Int) {
        if (depth == 0) {
            val left = TreeNode(`val`)
            val right = TreeNode(`val`)
            left.left = node.left
            right.right = node.right
            node.left = left
            node.right = right
        } else {
            if (node.left != null) {
                dfs(node.left!!, depth - 1, `val`)
            }
            if (node.right != null) {
                dfs(node.right!!, depth - 1, `val`)
            }
        }
    }
}

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