python3 解法, 执行用时: 40 ms, 内存消耗: 15.3 MB, 提交时间: 2022-05-30 10:59:22

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sumRootToLeaf(self, root: Optional[TreeNode]) -> int:
        def dfs(node: Optional[TreeNode], s: int) -> int:
            if node  is None:
                return 0
            
            s = (s << 1) | node.val
            if node.left == None and node.right == None:
                return s
            return dfs(node.left, s) + dfs(node.right, s)
        
        return dfs(root, 0)

golang 解法, 执行用时: 0 ms, 内存消耗: 2.9 MB, 提交时间: 2022-05-30 10:55:47

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func sumRootToLeaf(root *TreeNode) int {
    return helper(root, 0)
}

func helper(node *TreeNode, sum int) int {
    if node == nil {
        return 0
    }

    sum = sum << 1 + node.Val

    if node.Left == nil && node.Right == nil {
        return sum
    }

    return helper(node.Left, sum) + helper(node.Right, sum)
}