★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/9958594.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 a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3]
1
/
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:
Input: [4,9,0,5,1]
4
/
9 0
/
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。
例如，从根到叶子节点路径 1->2->3 代表数字 123。
计算从根到叶子节点生成的所有数字之和。
说明: 叶子节点是指没有子节点的节点。
示例 1:
输入: [1,2,3]
1
/
2 3
输出: 25
解释:
从根到叶子节点路径 1->2 代表数字 12.
从根到叶子节点路径 1->3 代表数字 13.
因此，数字总和 = 12 + 13 = 25.
示例 2:
输入: [4,9,0,5,1]
4
/
9 0
/
5 1
输出: 1026
解释:
从根到叶子节点路径 4->9->5 代表数字 495.
从根到叶子节点路径 4->9->1 代表数字 491.
从根到叶子节点路径 4->0 代表数字 40.
因此，数字总和 = 495 + 491 + 40 = 1026.
12ms
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 sumNumbers(_ root: TreeNode?) -> Int { 16 guard let root = root else { return 0 } 17 var sum = 0 18 var stack: [(node: TreeNode, accum: Int)] = [(node: root, accum: 0)] 19 20 while !stack.isEmpty { 21 let currentVal = stack.removeLast() 22 let updatedAccum = currentVal.accum * 10 + currentVal.node.val 23 24 guard currentVal.node.left != nil || currentVal.node.right != nil else { 25 sum += updatedAccum 26 continue 27 } 28 29 if let leftNode = currentVal.node.left { 30 stack.append((node: leftNode, accum: updatedAccum)) 31 } 32 33 if let rightNode = currentVal.node.right { 34 stack.append((node: rightNode, accum: updatedAccum)) 35 } 36 } 37 38 return sum 39 } 40 41 }
16ms
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 sumNumbers(_ root: TreeNode?) -> Int { 16 if root == nil { return 0 } 17 else { 18 let result = getNumber(root!, 0) 19 return result 20 } 21 } 22 23 func getNumber(_ node: TreeNode, _ number: Int) -> Int { 24 let currentNumber = number * 10 + node.val 25 var result = 0 26 if node.left == nil && node.right == nil { 27 return currentNumber 28 } else if node.left == nil { 29 result = getNumber(node.right!, currentNumber) 30 } else if node.right == nil { 31 result = getNumber(node.left!, currentNumber) 32 } else { 33 result = getNumber(node.left!, currentNumber) + getNumber(node.right!, currentNumber) 34 } 35 return result 36 } 37 }
20ms
1 class Solution { 2 func sumNumbers(_ root: TreeNode?) -> Int { 3 let nums = numbers(root) 4 print(nums) 5 return nums.reduce(0, { $0 + Int($1)! }) 6 } 7 8 func numbers(_ root: TreeNode?) -> [String] { 9 guard let root = root else { return [] } 10 11 if root.left == nil && root.right == nil { 12 return [String(root.val)] 13 } 14 15 return (numbers(root.left) + numbers(root.right)).map { String(root.val) + $0 } 16 } 17 }
24ms
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 sumNumbers(_ root: TreeNode?) -> Int { 16 var sum = 0 17 treeShowNodeVal(root: root, item: "", sum: &sum) 18 return sum 19 } 20 func treeShowNodeVal(root: TreeNode?, item: String, sum: inout Int){ 21 if root == nil { 22 return 23 } 24 var newItem = item 25 newItem.append(String((root?.val)!)) 26 if root?.left == nil && root?.right == nil { 27 let itemValue = Int(newItem) 28 sum = sum + itemValue! 29 } 30 treeShowNodeVal(root: root?.left, item: newItem, sum: &sum) 31 treeShowNodeVal(root: root?.right, item: newItem, sum: &sum) 32 } 33 }
28ms
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 sumNumbers(_ root: TreeNode?) -> Int { 16 var array = [String]() 17 guard let root = root else { return 0 } 18 19 func helper(_ node: TreeNode, _ temp: String) { 20 var temp = temp + "(node.val)" 21 if node.left == nil && node.right == nil { 22 array.append(temp) 23 } else { 24 if let left = node.left { 25 helper(left, temp) 26 } 27 if let right = node.right { 28 helper(right, temp) 29 } 30 } 31 } 32 33 helper(root, "") 34 return array.reduce(0, {return $0 + Int($1)!}) 35 } 36 }