[ LeetCode 590 ] N 叉树的后序遍历

原文连接：www.keketec.club/posts/11228…

天天分享一个LeetCode题目

天天 5 分钟，一块儿进步

LeetCode N 叉树的后序遍历，地址: leetcode-cn.com/problems/n-…

树结点类

class TreeNode(object):
    def __init__(self, val, children=[]):
        self.val = val
        self.children = children
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N 叉树的后序遍历

利用递归，依然遵循「左右根」的遍历原则

def postorder(self, root):
    if not root:
        return
    for node in root.children:
        self.postorder(node)
    print(root.val, end=" ")
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是否是看起来特别简单，可是这样不符合 LeetCode 题目中的要求

须要将结果放到一个数组中，因此须要提早初始化一个 list 进行存放

从新编码看看

def postorder_lc(self, root):
    res = []

    def post_order(root):
        if not root:
            return
        for node in root.children:
            post_order(node)
        res.append(root.val)

    post_order(root)
    return res
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就是提早初始化了 res，而后在遍历的时候进赋值操做

完整代码

可直接执行

# -*- coding:utf-8 -*-
# !/usr/bin/env python

# 树结点类
class Node(object):
    def __init__(self, val=None, children=[]):
        self.val = val
        self.children = children


class Solution(object):
    def postorder(self, root):
        if not root:
            return
        for node in root.children:
            self.postorder(node)
        print(root.val, end=" ")

    def postorder_lc(self, root):
        res = []

        def post_order(root):
            if not root:
                return
            for node in root.children:
                post_order(node)
            res.append(root.val)

        post_order(root)
        return res


if __name__ == "__main__":
    # 新建节点
    root = Node('A')
    node_B = Node('B')
    node_C = Node('C')
    node_D = Node('D')
    node_E = Node('E')
    node_F = Node('F')
    node_G = Node('G')
    node_H = Node('H')
    node_I = Node('I')
    # 构建三叉树
    # A
    # / | \
    # B C D
    # /|\ / \
    # E F G H I
    root.children = [node_B, node_C, node_D]
    node_B.children = [node_E, node_F, node_G]
    node_D.children = [node_H, node_I]

    s = Solution()
    s.postorder(root)
    print("\n")
    print(s.postorder_lc(root))
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