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Given a binary tree containing digits from 0-9
only, each root-to-leaf path could represent a number.node
An example is the root-to-leaf path 1->2->3
which represents the number 123
.git
Find the total sum of all root-to-leaf numbers.github
Note: A leaf is a node with no children.微信
Example:app
Input: [1,2,3] 1 / \ 2 3 Output: 25 Explanation: The root-to-leaf path represents the number . The root-to-leaf path represents the number . Therefore, sum = 12 + 13 = .1->2121->31325
Example 2:spa
Input: [4,9,0,5,1] 4 / \ 9 0 / \ 5 1 Output: 1026 Explanation: The root-to-leaf path represents the number 495. The root-to-leaf path represents the number 491. The root-to-leaf path represents the number 40. Therefore, sum = 495 + 491 + 40 = .4->9->54->9->14->01026
給定一個二叉樹,它的每一個結點都存放一個 0-9
的數字,每條從根到葉子節點的路徑都表明一個數字。code
例如,從根到葉子節點路徑 1->2->3
表明數字 123
。htm
計算從根到葉子節點生成的全部數字之和。blog
說明: 葉子節點是指沒有子節點的節點。
示例 1:
輸入: [1,2,3] 1 / \ 2 3 輸出: 25 解釋: 從根到葉子節點路徑 表明數字 . 從根到葉子節點路徑 表明數字 . 所以,數字總和 = 12 + 13 = .1->2121->31325
示例 2:
輸入: [4,9,0,5,1] 4 / \ 9 0 / \ 5 1 輸出: 1026 解釋: 從根到葉子節點路徑 表明數字 495. 從根到葉子節點路徑 表明數字 491. 從根到葉子節點路徑 表明數字 40. 所以,數字總和 = 495 + 491 + 40 = .4->9->54->9->14->01026
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 }