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: 25Explanation:
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: 1026Explanation:
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
}
}
