1022. Sum of Root To Leaf Binary Numbers

Easy

You are given the root of a binary tree where each node has a 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.

The test cases are generated so that the answer fits in a 32-bits integer.

Example 1:

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

Example 2:

Input: root = [0]
Output:
 0

Constraints:

• The number of nodes in the tree is in the range [1, 1000].

• Node.val is 0 or 1.

解題

一開始的笨方法，沒有使用 bitwise 來處理。

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sumRootToLeaf(root *TreeNode) int {
    sum := 0
    
    var recursive func(*TreeNode, string) 
    recursive = func(root *TreeNode, str string) {
        if root.Left==nil && root.Right==nil {
            sum += convert(str)
        }
        
        if root.Left!=nil {
            recursive(root.Left, str+strconv.Itoa(root.Left.Val))
        }
        
        if root.Right!=nil {
            recursive(root.Right, str+strconv.Itoa(root.Right.Val))
        }
    }
    
    recursive(root, strconv.Itoa(root.Val))
    // fmt.Println(sum)
    
    return sum
}

func convert(str string) int {
    if i, err := strconv.ParseInt(str, 2, 64); err != nil {
        fmt.Println(err)
    } else {
        return int(i)
    }
    
    return 0
}

參考了以下別人的寫法

func sumRootToLeaf(root *TreeNode) int {
    return dfs(root, 0)
}

func dfs(root *TreeNode, currSum int) int {
    currSum = (currSum << 1) | root.Val
    
    if root.Left == nil && root.Right == nil {
        return currSum
    }
    
    total := 0
    if root.Left != nil {
        total += dfs(root.Left, currSum)
    }
    
    if root.Right != nil {
        total += dfs(root.Right, currSum)
    }
    
    return total
}

我改良了一下原本的程式碼

Runtime: 0 ms, faster than 100.00%

Memory Usage: 3.1 MB, less than 54.76%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sumRootToLeaf(root *TreeNode) int {
    sum := 0
    
    var recursive func(*TreeNode, int) 
    recursive = func(root *TreeNode, currSum int) {
        currSum = (currSum << 1) | root.Val
        
        if root.Left==nil && root.Right==nil {
            sum += currSum
        }
        
        if root.Left!=nil {
            recursive(root.Left, currSum)
        }
        
        if root.Right!=nil {
            recursive(root.Right, currSum)
        }
    }
    
    recursive(root, 0)
    
    return sum
}