107. Binary Tree Level Order Traversal II

Easy

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

For example:
Given binary tree [3,9,20,null,null,15,7],

  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

1. Use queue to do level traversal + deque.appendleft or list[ : : -1] to realize bottom-up

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque
class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        res = deque()
        q = []
        q.append(root)
        while(q):
            temp = []
            l = len(q)
            for _ in range(l):
                cur = q.pop(0)
                temp.append(cur.val)
                if cur.left: q.append(cur.left)
                if cur.right: q.append(cur.right)
            res.appendleft(temp) # appendleft to realize bottom up
        return res

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        res = []
        q = []
        q.append(root)
        while(q):
            temp = []
            l = len(q)
            for _ in range(l):
                cur = q.pop(0)
                temp.append(cur.val)
                if cur.left: q.append(cur.left)
                if cur.right: q.append(cur.right)
            res.append(temp)
        return res[::-1] # list[::-1]

Time complexity is O(n)

2. DFS + record level

class Solution:
    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
        res = []
        self.dfs(res, 0, root)
        return res
    
    def dfs(self, res, level, node):
        if not node:
            return
        else:
            if len(res) < (level + 1):
                res.insert(0, [])  # notice: here insert is O(n)
            res[-(level+1)].append(node.val)
            self.dfs(res, level + 1, node.left)
            self.dfs(res, level + 1, node.right)

So time complexity is O(n^2)