107. Binary Tree Level Order Traversal II

Description

Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).

Example 1:

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

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].
• -1000 <= Node.val <= 1000

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 levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        result = []
        if not root:
            return result
        
        q = deque([root])

        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)
            result.append(level)

        return result[::-1]

Runtime: 0 ms | Beats 100.00%
Memory: 19.44 MB | Beats 72.06%

102. Binary Tree Level Order Traversal과 동일하고 마지막에 result의 원소 순서만 거꾸로 했다.

Other Solutions

1st

class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        list1=[]
        q=deque()
        q.append(root)
        while q:
            level=[]
            for i in range(len(q)):
                poping=q.popleft()
                if poping:
                    level.append(poping.val)
                    q.append(poping.left)
                    q.append(poping.right)
            if level:
                list1.append(level)
        return list1[::-1]

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