103_Binary Tree Zigzag Level Order Traversal
103. Binary Tree Zigzag Level Order Traversal
Level: medium
Tag: tree, breadth-first search
Question
Given a binary tree, return the zigzag level order traversal of its nodes' values.
(ie, from left to right, then right to left for the next level and alternate between).
Example 1
Input:
3
/ \
9 20
/ \
15 7
Output:
[
[3],
[20,9],
[15,7]
]
Idea (breadth-first search)
For each level, find all the node value, append to the result as a list. [reverse the node values if it's in even level]
Have a temporary list to store all the left and right child for all the node in the level.
Solution : breadth-first search
Time:
Space:
# 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 zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
if root == None:
return res
level = [root]
# use 1 and -1 to denote odd and even level
flag = 1
while len(level) > 0:
# For each level, find all the node value, append to the result as a list
res.append([node.val for node in level][::flag])
flag *= -1
# a temporary list to store all the left and right child for all the node in the level
temp = []
# append left and right child if they are not None
for node in level:
if node.left != None:
temp.append(node.left)
if node.right != None:
temp.append(node.right)
level = temp
return res
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