golang 解法, 执行用时: 4 ms, 内存消耗: 2.5 MB, 提交时间: 2020-11-18 11:45:39

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func binaryTreePaths(root *TreeNode) []string {
    if root == nil {
        return []string{}
    }
    if root.Left == nil && root.Right == nil {
        return []string{strconv.Itoa(root.Val)}
    }
    var paths []string

    if root.Left != nil {
        for _, i := range binaryTreePaths(root.Left) {
            paths = append(paths, strconv.Itoa(root.Val) + "->" + i)
        }
    }
    if root.Right != nil {
        for _, i := range binaryTreePaths(root.Right) {
            paths = append(paths, strconv.Itoa(root.Val) + "->" + i)
        }
    }

    return paths
}

python3 解法, 执行用时: 44 ms, 内存消耗: 13.7 MB, 提交时间: 2020-11-02 16:13:45

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        res = []
        if not root:
            return res
        stack = [(root, '')]
        while stack:
            node, road = stack.pop(0)
            if node.left == None and node.right == None:
                res.append(road + str(node.val))
            if node.left:
                stack.append((node.left, f'{road}{node.val}->'))
            if node.right:
                stack.append((node.right, f'{road}{node.val}->'))
        return res
            

python3 解法, 执行用时: 52 ms, 内存消耗: 13.5 MB, 提交时间: 2020-11-02 16:06:05

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        if not root:
            return []
        if not root.left and not root.right:
            return [str(root.val)]
        paths = []
        if root.left:
            for i in self.binaryTreePaths(root.left):
                paths.append(str(root.val) + '->' + i)
        if root.right:
            for i in self.binaryTreePaths(root.right):
                paths.append(str(root.val) + '->' + i)
        return paths