python3 解法, 执行用时: 44 ms, 内存消耗: 16.6 MB, 提交时间: 2022-11-14 10:32:25

# 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 pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
        ret = list()
        path = list()
        
        def dfs(root: TreeNode, targetSum: int):
            if not root:
                return
            path.append(root.val)
            targetSum -= root.val
            if not root.left and not root.right and targetSum == 0:
                ret.append(path[:])
            dfs(root.left, targetSum)
            dfs(root.right, targetSum)
            path.pop()
        
        dfs(root, targetSum)
        return ret

golang 解法, 执行用时: 8 ms, 内存消耗: 4.4 MB, 提交时间: 2022-11-14 10:31:59

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) (ans [][]int) {
    path := []int{}
    var dfs func(*TreeNode, int)
    dfs = func(node *TreeNode, left int) {
        if node == nil {
            return
        }
        left -= node.Val
        path = append(path, node.Val)
        defer func() { path = path[:len(path)-1] }()
        if node.Left == nil && node.Right == nil && left == 0 {
            ans = append(ans, append([]int(nil), path...))
            return
        }
        dfs(node.Left, left)
        dfs(node.Right, left)
    }
    dfs(root, targetSum)
    return
}