[ LeetCode 590 ] N 叉樹的後序遍歷

原文連接：www.keketec.club/posts/11228…

天天分享一個LeetCode題目

天天 5 分鐘，一塊兒進步

LeetCode N 叉樹的後序遍歷，地址: leetcode-cn.com/problems/n-…

樹結點類

class TreeNode(object):
    def __init__(self, val, children=[]):
        self.val = val
        self.children = children
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N 叉樹的後序遍歷

利用遞歸，依然遵循「左右根」的遍歷原則

def postorder(self, root):
    if not root:
        return
    for node in root.children:
        self.postorder(node)
    print(root.val, end=" ")
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是否是看起來特別簡單，可是這樣不符合 LeetCode 題目中的要求

須要將結果放到一個數組中，因此須要提早初始化一個 list 進行存放

從新編碼看看

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，而後在遍歷的時候進賦值操做

完整代碼

可直接執行

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

# 樹結點類
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__":
    # 新建節點
    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')
    # 構建三叉樹
    # 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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