-
Notifications
You must be signed in to change notification settings - Fork 0
/
112.path-sum.0.kt
58 lines (56 loc) · 1.45 KB
/
112.path-sum.0.kt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
/*
* @lc app=leetcode id=112 lang=kotlin
*
* [112] Path Sum
*
* https://leetcode.com/problems/path-sum/description/
*
* algorithms
* Easy (37.32%)
* Total Accepted: 298.5K
* Total Submissions: 798.9K
* Testcase Example: '[5,4,8,11,null,13,4,7,2,null,null,null,1]\n22'
*
* Given a binary tree and a sum, determine if the tree has a root-to-leaf path
* such that adding up all the values along the path equals the given sum.
*
* Note: A leaf is a node with no children.
*
* Example:
*
* Given the below binary tree and sum = 22,
*
*
* 5
* / \
* 4 8
* / / \
* 11 13 4
* / \ \
* 7 2 1
*
*
* return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
*
*/
/**
* 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_hasPathSum_0 {
fun hasPathSum(root: TreeNode?, sum: Int): Boolean =
getPaths(root).map { it.reduce { a, c -> a + c } }.any { it == sum }
fun getPaths(node: TreeNode?): List<List<Int>> = when {
node == null -> listOf<List<Int>>()
node.left == null && node.right == null -> listOf(listOf(node.`val`))
else -> (listOf(node.left, node.right)
.flatMap { getPaths(it) }
.map { listOf(node.`val`) + it })
}
}