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Solution.py
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Solution.py
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"""
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).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
"""
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
from collections import deque
from collections import defaultdict
class Solution(object):
def zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
level = 0
queue = deque([(root,level)])
ans_table = defaultdict(list)
while queue:
cur,level = queue.popleft()
ans_table[level].append(cur.val)
if cur.left:
queue.append((cur.left,level+1))
if cur.right:
queue.append((cur.right,level+1))
return [ans_table[i] if i % 2 == 0 else list(reversed(ans_table[i])) for i in range(len(ans_table)) ]
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
from collections import deque
from collections import defaultdict
queue = deque([])
level_map = defaultdict(list)
queue.append([root, 0])
while queue:
root, level = queue.popleft()
level_map[level].append(root.val)
if root.left:
queue.append([root.left, level + 1])
if root.right:
queue.append([root.right, level + 1])
return [ v if k % 2 == 0 else v[::-1] for k,v in level_map.items()]