[leetcode] 102. Binary Tree Level Order Traversal @ python

二叉树层次遍历

最新推荐文章于 2020-01-03 13:57:10 发布

原创最新推荐文章于 2020-01-03 13:57:10 发布 · 92 阅读

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本文介绍了二叉树的层次遍历算法，提供了两种宽度优先搜索的实现方式，包括详细的Python代码示例，展示了如何从左到右，按层获取二叉树节点的值。

原题

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

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

解法1

宽度优先搜索, 每次用列表储存当前层的节点, 按层进行遍历, 将结果加入到ans里.然后列表更新, 储存下一层的节点.
Time: O(V), V是二叉树中节点的数量
Space: O(n)

代码

# 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):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        # edge case
        if not root: return []
        ans = []
        q = [root]
        while q:
            ans.append([node.val for node in q])
            new_q = []
            for node in q:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            q = new_q 
            
        return ans

解法2

解法1的简化版, 使用list comprehension, 直接得到下一层的节点. 注意如果root为空时, q为[None], 程序会报错, 因此while循环的条件是root和q都不为空.

Time: O(V), V是二叉树中节点的数量
Space: O(n)

代码

# 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):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        ans, q = [], [root]       
        while root and q:
            ans.append([node.val for node in q])
            q = [kid for node in q for kid in (node.left, node.right) if kid]
            
        return ans