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[LeetCode]102. 二叉树的层次遍历（Binary Tree Level Order Traversal）

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题目描述

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

例如:
给定二叉树: [3,9,20,null,null,15,7],

3
/ \
9 20
/ \
15 7

返回其层次遍历结果：

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

解题思路

  1. 解法一：递归。DFS。
  2. 解法二：迭代。BFS。

代码

解法一：递归

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

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        levels = []
        if root is None:
            return levels
        
        def helper(node, level):
            if level == len(levels):
                levels.append([])
            levels[level].append(node.val)
            if node.left is not None:
                helper(node.left, level + 1)
            if node.right is not None:
                helper(node.right, level + 1)
        
        helper(root, 0)
        return levels

执行用时 : 100 ms
内存消耗 : 14.7 MB

解法二：迭代

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

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        levels = []
        if root is None:
            return levels
        
        queue = [root]
        level = 0
        while queue:
            levels.append([])
            length = len(queue)
            for _ in range(length):
                node = queue.pop(0)
                levels[level].append(node.val)
                if node.left is not None:
                    queue.append(node.left)
                if node.right is not None:
                    queue.append(node.right)
            level += 1
        return levels

执行用时 : 56 ms
内存消耗 : 14 MB

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