129. Sum Root to Leaf Numbers
Medium 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 3Output: 25复制代码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 1Output: 1026复制代码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.
题目大意： 将一棵树从root -> leaf的一个遍历当成一个整数；求各条 root -> leaf遍历的和; 简单的树的遍历；下面给出两种思路：
- 递归
- 循环
代码如下(Swift5):
class Solution { // 递归 // Runtime: 8 ms, faster than 100.00% of Swift online submissions for Sum Root to Leaf Numbers. // Memory Usage: 19.1 MB, less than 25.00% of Swift online submissions for Sum Root to Leaf Numbers. func sumNumbers(_ root: TreeNode?) -> Int { var result = 0 func next(_ previous: Int, node: TreeNode?) { guard let node = node else { return } let current = previous * 10 + node.val if node.left == nil && node.right == nil { result += current } else { next(current, node: node.left) next(current, node: node.right) } } next(0, node: root) return result } // 循环 // Runtime: 12 ms, faster than 89.86% of Swift online submissions for Sum Root to Leaf Numbers. // Memory Usage: 18.9 MB, less than 50.00% of Swift online submissions for Sum Root to Leaf Numbers. func sumNumbers_2(_ root: TreeNode?) -> Int { typealias keyNode = (previous: Int, node: TreeNode) guard let root = root else { return 0 } var result = 0 var keyNodes: [keyNode] = [(0, root)] while !keyNodes.isEmpty { var temp: [keyNode] = [] for keyNode in keyNodes { let node = keyNode.node let current = keyNode.previous * 10 + node.val if node.left == nil, node.right == nil { result += current } else { if let left = node.left { temp.append((current, left)) } if let right = node.right { temp.append((current, right)) } } } keyNodes = temp } return result }}复制代码