[Swift]LeetCode1022. 从根到叶的二进制数之和 | Sum of Root To Leaf Binary Numbers

Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path represents a binary number starting with the most significant bit.  For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

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Return the sum of these numbers modulo 10^9 + 7.

Example 1:

Input: [1,0,1,0,1,0,1]
Output: 22 Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22 

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.

给出一棵二叉树，其上每个结点的值都是 0 或 1 。每一条从根到叶的路径都代表一个从最高有效位开始的二进制数。例如，如果路径为 0 -> 1 -> 1 -> 0 -> 1，那么它表示二进制数 01101，也就是 13 。

对树上的每一片叶子，我们都要找出从根到该叶子的路径所表示的数字。

以 10^9 + 7 为模，返回这些数字之和。 

示例：

输入：[1,0,1,0,1,0,1]
输出：22
解释：(100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22 

提示：

1. 树中的结点数介于 1 和 1000 之间。
2. node.val 为 0 或 1 。

Runtime: 24 ms Memory Usage: 19 MB

 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     var mod:Int = 1000000007
16     var ans:Int = 0
17     func sumRootToLeaf(_ root: TreeNode?) -> Int {
18         ans = 0
19         dfs(root, 0)
20         return ans % mod
21     }
22     
23     func dfs(_ cur: TreeNode?,_ v:Int)
24     {
25         if cur == nil {return}
26         if cur?.left == nil && cur?.right == nil
27         {
28             ans += v*2 + cur!.val
29             return
30         }
31         dfs(cur?.left, (v*2 + cur!.val) % mod)
32         dfs(cur?.right, (v*2 + cur!.val) % mod)
33         
34     }
35 }

 

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