https://leetcode.com/problems/binary-tree-level-order-traversal/
Binary Tree Level Order Traversal
Total Accepted: 70821 Total Submissions: 237732 Difficulty: Easy
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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Code
# 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]]
"""
res = []
self.dfs(root, 0, res)
return res
def dfs(self, root, level, res):
if root:
if len(res)<level+1:
res.append([])
res[level].append(root.val)
self.dfs(root.left, level+1, res)
self.dfs(root.right, level+1, res)
Idea
Similar to previous post.