diff --git a/work/1/Medium/Tree/102.binary-tree-level-order-traversal.py b/work/1/Medium/Tree/102.binary-tree-level-order-traversal.py new file mode 100644 index 0000000..075c31f --- /dev/null +++ b/work/1/Medium/Tree/102.binary-tree-level-order-traversal.py @@ -0,0 +1,66 @@ +""" +102. Binary Tree Level Order Traversal +Difficulty: Medium +https://leetcode.com/problems/binary-tree-level-order-traversal/ + +────────────────────────────────────────────────── + +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 +""" + +# 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 +from collections import deque + + +class Solution: + # T: O(n) S: O(n) + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + res = [] + + q = deque() + q.append(root) + + while q: + qLen = len(q) + level = [] + for i in range(qLen): + node = q.popleft() + if node: + level.append(node.val) + q.append(node.left) + q.append(node.right) + if level: + res.append(level) + + return res