[ LeetCode 589 ] N 叉树的前序遍历

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LeetCode N 叉树的前序遍历，地址: leetcode-cn.com/problems/n-…数组

树结点类

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

利用递归，依然遵循「跟左右」的遍历原则markdown

def preorder(self, root):
    print(root.val, end=" ")
    for node in root.children:
        self.preorder(node)
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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app

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oop

从新编码看看ui

def preorder(self, root):
    # print(root.val, end=" ")
    # for node in root.children:
    # self.preorder(node)

    res = []
    def dfs(root):
        if not root:
            return
        res.append(root.val)
        for node in root.children:
            dfs(node)
    dfs(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编码

完整代码

可直接执行spa

# -*- 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 preorder(self, root):
        # print(root.val, end=" ")
        # for node in root.children:
        # self.preorder(node)

        res = []
        def dfs(root):
            if not root:
                return
            res.append(root.val)
            for node in root.children:
                dfs(node)
        dfs(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()
    print(s.preorder(root))
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