[Swift]LeetCode1161. 最大層內元素和 | Maximum Level Sum of a Binary Tree

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：爲敢（WeiGanTechnologies）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-gnqfvfoj-km.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.

Return the smallest level X such that the sum of all the values of nodes at level X is maximal.

Example 1:

Input: [1,7,0,7,-8,null,null]
Output: 2 Explanation: Level 1 sum = 1. Level 2 sum = 7 + 0 = 7. Level 3 sum = 7 + -8 = -1. So we return the level with the maximum sum which is level 2.

Note:

1. The number of nodes in the given tree is between 1 and 10^4.
2. -10^5 <= node.val <= 10^5

---

給你一個二叉樹的根節點 root。設根節點位於二叉樹的第 1 層，而根節點的子節點位於第 2 層，依此類推。

請你找出層內元素之和 最大 的那幾層（可能只有一層）的層號，並返回其中 最小 的那個。

示例：

輸入：[1,7,0,7,-8,null,null]
輸出：2
解釋：
第 1 層各元素之和爲 1，
第 2 層各元素之和爲 7 + 0 = 7，
第 3 層各元素之和爲 7 + -8 = -1，
因此咱們返回第 2 層的層號，它的層內元素之和最大。

提示：

1. 樹中的節點數介於 1 和 10^4 之間
2. -10^5 <= node.val <= 10^5

---

740ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     var levels: [Int] = []
16     func maxLevelSum(_ root: TreeNode?) -> Int {
17         traverse(root, 0)
18         var maxValue = root!.val
19         var maxValueIndex = 0
20         for (index, level) in levels.enumerated() {
21             if level > maxValue {
22                 maxValue = level
23                 maxValueIndex = index
24             }
25         }
26         return maxValueIndex + 1
27     }
28     
29     func traverse(_ node: TreeNode?, _ level: Int) {
30         if let node = node {
31             if level >= levels.count {
32                 levels.append(0)
33             }
34             levels[level] += node.val
35             traverse(node.left, level + 1)
36             traverse(node.right, level + 1)
37         }
38     }
39 }

---

764ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func maxLevelSum(_ root: TreeNode?) -> Int {
16         var levels: [Int: Int] = [:]
17         dfs(node: root, depth: 1, levels: &levels)
18         var result = 0
19         var maximum = levels[0] ?? Int.min
20         for level in levels {
21             if maximum < (level.value ?? Int.min) {
22                 // new minimum
23                 maximum = (level.value ?? Int.min)
24                 result = level.key
25             }
26         }
27         
28         return result
29     }
30     
31     private func dfs(node: TreeNode?, depth: Int, levels: inout [Int: Int]) {
32         guard let node = node else { return }
33         
34         levels[depth] = (levels[depth] ?? 0) + node.val
35         
36         dfs(node: node.left, depth: depth + 1, levels: &levels)
37         dfs(node: node.right, depth: depth + 1, levels: &levels)
38     }
39 }

---

808ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15   var maxs = [Int](repeating: 0, count: 10001)
16   func maxLevelSum(_ root: TreeNode?) -> Int {
17     helper(root, 1)
18     return maxs.enumerated().max { (a, b) -> Bool in
19       a.element <= b.element
20     }!.offset
21   }
22   
23   func helper(_ node: TreeNode?, _ lvl: Int) {
24     guard let n = node else {
25       return
26     }
27     maxs[lvl] += n.val
28     helper(n.right, lvl + 1)
29     helper(n.left, lvl + 1)
30   }
31 }

---

Runtime: 828 ms

Memory Usage: 21 MB

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     var s:[Int:Int] = [Int:Int]()
16     func maxLevelSum(_ root: TreeNode?) -> Int {
17         dfs(root,1)
18         var best:Int = s[1,default:0]
19         var v:Int = 1
20         for (key,val) in s
21         {
22             if val > best
23             {
24                 best = val
25                 v = key
26             }            
27         }
28         return v
29     }
30     
31     func dfs(_ root: TreeNode?,_ level:Int)
32     {
33         if root == nil {return}
34         dfs(root!.left, level + 1)
35         dfs(root!.right, level + 1)
36         s[level,default:0] += root!.val
37     }
38 }