[Swift]LeetCode1161. 最大层内元素和 | Maximum Level Sum of a Binary Tree

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（www.zengqiang.org）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/11371957.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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

---

给你一个二叉树的根节点 root。设根节点位于二叉树的第 1 层，而根节点的子节点位于第 2 层，依此类推。

请你找出层内元素之和 最大 的那几层（可能只有一层）的层号，并返回其中 最小 的那个。

示例：

输入：[1,7,0,7,-8,null,null]
输出：2
解释：
第 1 层各元素之和为 1，
第 2 层各元素之和为 7 + 0 = 7，
第 3 层各元素之和为 7 + -8 = -1，
所以我们返回第 2 层的层号，它的层内元素之和最大。

提示：

1. 树中的节点数介于 1 和 10^4 之间
2. -10^5 <= node.val <= 10^5

---

740ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    var levels: [Int] = []
    func maxLevelSum(_ root: TreeNode?) -> Int {
        traverse(root, 0)
        var maxValue = root!.val
        var maxValueIndex = 0
        for (index, level) in levels.enumerated() {
            if level > maxValue {
                maxValue = level
                maxValueIndex = index
            }
        }
        return maxValueIndex + 1
    }
    
    func traverse(_ node: TreeNode?, _ level: Int) {
        if let node = node {
            if level >= levels.count {
                levels.append(0)
            }
            levels[level] += node.val
            traverse(node.left, level + 1)
            traverse(node.right, level + 1)
        }
    }
}

---

764ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func maxLevelSum(_ root: TreeNode?) -> Int {
        var levels: [Int: Int] = [:]
        dfs(node: root, depth: 1, levels: &levels)
        var result = 0
        var maximum = levels[0] ?? Int.min
        for level in levels {
            if maximum < (level.value ?? Int.min) {
                // new minimum
                maximum = (level.value ?? Int.min)
                result = level.key
            }
        }
        
        return result
    }
    
    private func dfs(node: TreeNode?, depth: Int, levels: inout [Int: Int]) {
        guard let node = node else { return }
        
        levels[depth] = (levels[depth] ?? 0) + node.val
        
        dfs(node: node.left, depth: depth + 1, levels: &levels)
        dfs(node: node.right, depth: depth + 1, levels: &levels)
    }
}

---

808ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
  var maxs = [Int](repeating: 0, count: 10001)
  func maxLevelSum(_ root: TreeNode?) -> Int {
    helper(root, 1)
    return maxs.enumerated().max { (a, b) -> Bool in
      a.element <= b.element
    }!.offset
  }
  
  func helper(_ node: TreeNode?, _ lvl: Int) {
    guard let n = node else {
      return
    }
    maxs[lvl] += n.val
    helper(n.right, lvl + 1)
    helper(n.left, lvl + 1)
  }
}

---

Runtime: 828 ms

Memory Usage: 21 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    var s:[Int:Int] = [Int:Int]()
    func maxLevelSum(_ root: TreeNode?) -> Int {
        dfs(root,1)
        var best:Int = s[1,default:0]
        var v:Int = 1
        for (key,val) in s
        {
            if val > best
            {
                best = val
                v = key
            }            
        }
        return v
    }
    
    func dfs(_ root: TreeNode?,_ level:Int)
    {
        if root == nil {return}
        dfs(root!.left, level + 1)
        dfs(root!.right, level + 1)
        s[level,default:0] += root!.val
    }
}