LeetCode in Kotlin

1022. Sum of Root To Leaf Binary Numbers

Easy

You are given the root of a binary tree where each node has a value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit.

• For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return the sum of these numbers.

The test cases are generated so that the answer fits in a 32-bits integer.

Example 1:

Input: root = [1,0,1,0,1,0,1]

Output: 22

Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Example 2:

Input: root = [0]

Output: 0

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• Node.val is 0 or 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 sumRootToLeaf(root: TreeNode?): Int {
        val paths: MutableList<List<Int>> = ArrayList()
        dfs(root, paths, ArrayList())
        var sum = 0
        for (list in paths) {
            var num = 0
            for (i in list) {
                num = (num shl 1) + i
            }
            sum += num
        }
        return sum
    }

    private fun dfs(root: TreeNode?, paths: MutableList<List<Int>>, path: MutableList<Int>) {
        path.add(root!!.`val`)
        if (root.left != null) {
            dfs(root.left!!, paths, path)
            path.removeAt(path.size - 1)
        }
        if (root.right != null) {
            dfs(root.right!!, paths, path)
            path.removeAt(path.size - 1)
        }
        if (root.left == null && root.right == null) {
            paths.add(ArrayList(path))
        }
    }
}

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