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] ]
Solution:
# 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 levelOrder(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ q, res = root and [root], [] while q: size, level_path = len(q), [] for i in range(size): node = q.pop(0) level_path.append(node.val) if node.left: q.append(node.left) if node.right: q.append(node.right) res.append(level_path) return res
最经典的队列实现BFS,注意size是作为for循环次数的依据,Python的 for node in q 的循环不适用,因为q在动态变化。
N-ary的树可参考559(https://www.cnblogs.com/lowkeysingsing/p/11152751.html)
原文地址:https://www.cnblogs.com/lowkeysingsing/p/11172103.html
时间: 2024-12-09 07:37:28