golang 解法, 执行用时: 0 ms, 内存消耗: 3.8 MB, 提交时间: 2024-02-19 10:17:05

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func postorder(root *Node) (ans []int) {
    var dfs func(node *Node)
    dfs = func(node *Node) {
        if node != nil {
            for _, child := range node.Children {
                dfs(child)
            }
            ans = append(ans, node.Val)
        }
    }
    dfs(root)
    return
}

python3 解法, 执行用时: 55 ms, 内存消耗: 18.1 MB, 提交时间: 2024-02-19 10:16:43

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        stack, ans = [root], []
        while stack:
            if (obj := stack.pop()) is not None:
                if type(obj) == Node:
                    stack += [obj.val] + obj.children[::-1]
                else:
                    ans.append(obj)
        return ans

python3 解法, 执行用时: 46 ms, 内存消耗: 18.1 MB, 提交时间: 2024-02-19 10:15:43

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""

class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        result = []
        def post_order(root):
            if root:
                for node in root.children:
                    post_order(node)
                result.append(root.val)
        post_order(root)
        return result

golang 解法, 执行用时: 0 ms, 内存消耗: 3.9 MB, 提交时间: 2021-06-17 10:44:35

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func postorder(root *Node) []int {
    var res = []int{}
    if root == nil{
        return res
    }
	var st = []*Node{root}
    for len(st) > 0 {
        x := st[len(st)-1]
        st = st[:len(st)-1]
        if x != nil {
            res = append(res, x.Val)
            st = append(st, x.Children...)
        }

    }
    n := len(res)
    for i := 0; i < n/2; i++ {
        res[i], res[n-1-i] = res[n-1-i], res[i]
    }
    return res
}