Quartz 4

binary tree level order traversal

2 min read

my solution - bfs

# 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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if root is None: return []
        
        q = deque([root])
        res = []
 
        while q:
            sublist = []
            for i in range(len(q)): # pop everything from that level
                cur = q.popleft()
                sublist.append(cur.val)
                if cur.left: q.append(cur.left)
                if cur.right: q.append(cur.right)
            res.append(sublist)
        
        return res
  • for some reasons this was easier than the easy questions??
  • edge case
    • the root might be null!! → we have to remember this
  • time complexity
    • O(N)
  • space complexity
    • O(N)
    • determined the maximum number of node in queue q at any single moment. the queue holds one complete level
    • so at the last level, it will hold approximately N/2 nodes → this is O(N/2) which simpliies to O(N)

apparently there is a dfs solution

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        res = []
 
        def dfs(node, depth):
            if not node:
                return None
            if len(res) == depth:
                res.append([])
 
            res[depth].append(node.val)
            dfs(node.left, depth + 1)
            dfs(node.right, depth + 1)
 
        dfs(root, 0)
        return res
  • umm…wtf is this 😭(this solution messed me up but after long staring i think i got it lmao)
  • basically… we calculate the “depth” of each node and use that value as index for res

do a quick run through with:

    3      (Level 0)
   / \
  9   20   (Level 1)

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