107. Binary Tree Level Order Traversal II [easy] (Python)

2023年1月15日 222次阅读 来源: coder_orz

题目链接

https://leetcode.com/problems/binary-tree-level-order-traversal-ii/

题目原文

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \   9  20
    /  \    15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

题目翻译

给定一个二叉树，返回其从下到上的层序遍历（从左到右，从下到上）。

思路方法

这道题实际上完全可以先得到从上到下的层序遍历，再逆序结果，但这样就没意思了，所以应该尝试避免reverse操作。不过方法还是类似的。
相关问题：102. Binary Tree Level Order Traversal [easy] (Python)

思路一

用队列来实现BFS，注意要在广搜的时候记录当前的层数。

代码

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

class Solution(object):
    def levelOrderBottom(self, root):
        """ :type root: TreeNode :rtype: List[List[int]] """
        res = []
        queue = [(root, 0)]
        while len(queue) > 0:
            node, depth = queue.pop()
            if node:
                if len(res) <= depth:
                    res.insert(0, [])
                res[-(depth+1)].append(node.val)
                queue.insert(0, (node.left, depth+1))
                queue.insert(0, (node.right, depth+1))
        return res

思路二

用栈来实现DFS，注意要在深搜的时候记录当前的深度。

代码

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

class Solution(object):
    def levelOrderBottom(self, root):
        """ :type root: TreeNode :rtype: List[List[int]] """
        res = []
        stack = [(root, 0)]
        while len(stack) > 0:
            node, depth = stack.pop()
            if node:
                if len(res) <= depth:
                    res.insert(0, [])
                res[-(depth+1)].append(node.val)
                stack.append((node.right, depth+1))
                stack.append((node.left, depth+1))
        return res

思路三

递归法实现DFS。

代码

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

class Solution(object):
    def levelOrderBottom(self, root):
        """ :type root: TreeNode :rtype: List[List[int]] """
        res = []
        self.dfs(root, 0, res)
        return res

    def dfs(self, root, depth, res):
        if root:
            if depth >= len(res):
                res.insert(0, [])
            res[-(depth+1)].append(root.val)
            self.dfs(root.left, depth+1, res)
            self.dfs(root.right, depth+1, res)

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！
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    原文作者：coder_orz
    原文地址: https://blog.csdn.net/coder_orz/article/details/51583729
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