102. Binary Tree Level Order Traversal

Medium LeetCode

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

Example 1

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

Output: [[3],[9,20],[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]
• -1000 <= Node.val <= 1000

How to solve the problem

Code

• BFS(Two Array)

# 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]]:
        # BFS with two array
        if root is None:
            return []
        ans = []
        cur = [root]
        while cur:
            nxt = []
            vals = []
            for node in cur:
                vals.append(node.val)
                if node.left:
                    nxt.append(node.left)
                if node.right:
                    nxt.append(node.right)
            cur = nxt
            ans.append(vals)
        return ans

• BFS(Queue)

# 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]]:
         # BFS with queue
        if root is None:
            return []
        ans = []
        q = deque([root])
        while q:
            vals = []
            for _ in range(len(q)):
                node = q.popleft()
                vals.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(vals)
        return ans

• BFS(Queue) Written by myself

# 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]]:
        # BFS with queue
        if root is None:
            return []
        result = []
        queue = deque([root]) # deque's object must be iterable
        while queue: # When queue is not empty, do the while loop
            level = []
            for _ in range(len(queue)):
                node = queue.popleft()
                level.append(node.val) # node structure: node = TreeNode(val=3, left=..., right=...)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            result.append(level)
        return result

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)