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

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10667991.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    var mod:Int = 1000000007
    var ans:Int = 0
    func sumRootToLeaf(_ root: TreeNode?) -> Int {
        ans = 0
        dfs(root, 0)
        return ans % mod
    }
    
    func dfs(_ cur: TreeNode?,_ v:Int)
    {
        if cur == nil {return}
        if cur?.left == nil && cur?.right == nil
        {
            ans += v*2 + cur!.val
            return
        }
        dfs(cur?.left, (v*2 + cur!.val) % mod)
        dfs(cur?.right, (v*2 + cur!.val) % mod)
        
    }
}

---

36ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    let mode = 1000_000_007

    func sumRootToLeaf(_ root: TreeNode?) -> Int {
        guard let node = root else{
            return 0
        }
        var answer = 0
        deepFirst(&answer, node.val, v: node)
        return answer
    }

    func deepFirst(_ answer: inout Int,_ current: Int, v : TreeNode){
        if v.left == nil && v.right == nil {
            answer = (answer + current) % mode
            return
        }
        if let lhs = v.left{
            deepFirst(&answer, (current * 2 + lhs.val)%mode, v: lhs)
        }
        if let rhs = v.right{
            deepFirst(&answer, (current * 2 + rhs.val)%mode, v: rhs)
        }
    }    
}

---

40ms

class Solution {
  private let module = 1_000_000_007
  
  func sumRootToLeaf(_ root: TreeNode?) -> Int {
    return helper(root: root, val: 0) % module
  }

  private func helper(root: TreeNode?, val: Int) -> Int {
    guard let node = root else {
      return 0
    }
    if node.left == nil && node.right == nil {
      return (val << 1 + node.val) % module
    }
    return helper(root: node.right, val: (val << 1 + node.val) % module) + 
      helper(root: node.left, val: (val << 1 + node.val) % module)
  }
}

---

44ms

class Solution {
    func sumRootToLeaf(_ root: TreeNode?) -> Int {
           return sumUp(root, 0)
    }

    func sumUp(_ node: TreeNode?, _ prefix: Int) -> Int {
        guard let node = node else {
            return prefix
        }
        let modNum = Int(pow(Double(10), Double(9))) + 7
        var prefix = ((prefix % modNum) * 2) % modNum + node.val
        var sum = 0
        var flag = false
        if let left = node.left {
            sum = (sum + sumUp(left, prefix)) % modNum
            flag = true
        }
        if let right = node.right {
            sum = (sum + sumUp(right, prefix)) % modNum
            flag = true
        }
        if !flag {
            return prefix
        }
        return sum % modNum
    }
}

---

48ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    var sum: Int = 0
    func sumRootToLeaf(_ root: TreeNode?) -> Int {
        guard let root = root else { return 0 }
        dfs(root, []) 
        return sum
    }
    
    func dfs(_ node: TreeNode, _ str: [Int]) {
        guard node.left != nil || node.right != nil else {
            sum = (sum + convert(str + [node.val])) % 1000000007
            return
        }
        
        if let left = node.left {
            dfs(left, str + [node.val])
        }
        if let right = node.right {
            dfs(right, str + [node.val])
        }
    }
        
    func convert(_ str: [Int]) -> Int {
        var num: Int = 0
        
        for n in str {
            num = (num * 2 + n) % 1000000007
        }
        
        return num
    }
}