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86 lines
2.1 KiB
Plaintext
86 lines
2.1 KiB
Plaintext
---
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title: '102. Binary Tree Level Order Traversal'
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description: 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)
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sidebar:
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label: 'Binary Tree Level Order Traversal'
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badge: 'Medium'
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---
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<Badge variant="accent">Tree</Badge>
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### Example 1:
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- Input: `root = [3,9,20,null,null,15,7]`
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- Output: `[[3],[9,20],[15,7]]`
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### Example 2:
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- Input: `root = [1]`
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- Output: `[[1]]`
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### Example 3:
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- Input: `root = []`
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- Output: `[]`
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### Constraints:
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- The number of nodes in the tree is in the range [0, 2000].
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- `-1000 <= Node.val <= 1000`
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## Approach
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```mermaid
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flowchart TD
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S(["levelOrder(root)"]) --> I["res = [], q = deque(), q.append(root)"]
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I --> W{"q non-empty?"}
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W -- no --> E(["return res"])
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W -- yes --> L["qLen = len(q), level = [] — qLen freezes the current level's size"]
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L --> F{"more i in range(qLen)?"}
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F -- no --> V{"level non-empty?"}
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V -- yes --> A["res.append(level)"]
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A --> W
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V -- no --> W
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F -- yes --> P["node = q.popleft()"]
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P --> N{"node is not None?"}
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N -- no --> F
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N -- yes --> C["level.append(node.val), q.append(node.left), q.append(node.right) — None children are filtered on the next level"]
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C --> F
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```
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## Solution
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```py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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from collections import deque
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class Solution:
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# T: O(n) S: O(n)
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def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
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res = []
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q = deque()
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q.append(root)
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while q:
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qLen = len(q)
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level = []
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for i in range(qLen):
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node = q.popleft()
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if node:
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level.append(node.val)
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q.append(node.left)
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q.append(node.right)
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if level:
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res.append(level)
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return res
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```
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## Explanation
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<YouTube id="6ZnyEApgFYg" title="Binary Tree Level Order Traversal - BFS - Leetcode 102" />
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