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 3 Output: 25 複製程式碼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 複製程式碼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.
題目大意：
將一棵樹從root -> leaf的一個遍歷當成一個整數；求各條 root -> leaf遍歷的和;
簡單的樹的遍歷；下面給出兩種思路：
- 遞迴
- 迴圈
程式碼如下(Swift5):
class Solution {
// 遞迴
// 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
}
// 迴圈
// 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
}
}
複製程式碼