103. Binary Tree Zigzag Level Order Traversal

Description

Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).

Example 1:

• Input: root = [3,9,20,null,null,15,7]
• Output: [[3],[20,9],[15,7]]

Example 2:

• Input: root = [1]
• Output: [[1]]

Example 3:

• Input: root = []
• Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Submitted Code

# 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: Optional[TreeNode]) -> List[List[int]]:
        result = []
        if not root:
            return result

        q = deque()
        q.append(root)
        is_reversed = 0

        while q:
            level = []

            for _ in range(len(q)):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
                level.append(node.val)

            if is_reversed:
                level.reverse()

            result.append(level)
            is_reversed ^= 1

        return result

Runtime: 0 ms | Beats 100.00%
Memory: 19.15 MB | Beats 99.52%

is_reversed로 각 level마다 정방향인지 역방향인지 체크하고 result에 넣기 전에 순서를 뒤집었다.

Other Solutions

1st

from collections import deque

class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root: return []
        res, dq, reverse = [], deque([root]), False

        while dq:
            level = []
            for _ in range(len(dq)):
                if not reverse:
                    node = dq.popleft()
                    level.append(node.val)
                    if node.left: dq.append(node.left)
                    if node.right: dq.append(node.right)
                else:
                    node = dq.pop()
                    level.append(node.val)
                    if node.right: dq.appendleft(node.right)
                    if node.left: dq.appendleft(node.left)
            res.append(level)
            reverse = not reverse
        return res

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑤)

이 방법은 deque의 양쪽을 모두 사용하기 때문에 자식을 넣는 순서가 달라진다.

2nd

class Solution:
    def zigzagLevelOrder(self, root):
        if not root: return []
        queue = deque([root])
        result, direction = [], 1
        
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:  queue.append(node.left)
                if node.right: queue.append(node.right)
            result.append(level[::direction])
            direction *= (-1)
        return result