102.二叉树的层次遍历.py

#
# @lc app=leetcode.cn id=102 lang=python3
#
# [102] 二叉树的层次遍历
#
# https://leetcode-cn.com/problems/binary-tree-level-order-traversal/description/
#
# algorithms
# Medium (60.80%)
# Likes:    402
# Dislikes: 0
# Total Accepted:    88.3K
# Total Submissions: 144.6K
# Testcase Example:  '[3,9,20,null,null,15,7]'
#
# 给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
# 
# 例如:
# 给定二叉树: [3,9,20,null,null,15,7],
# 
# ⁠   3
# ⁠  / \
# ⁠ 9  20
# ⁠   /  \
# ⁠  15   7
# 
# 
# 返回其层次遍历结果：
# 
# [
# ⁠ [3],
# ⁠ [9,20],
# ⁠ [15,7]
# ]
# 
# 
#

# @lc code=start
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        # BFS 队列
        if not root:
            return []
        res = []
        q = [root]
        while q:
            level = []
            for i in range(len(q)):
                cur = q.pop(0)
                level.append(cur.val)
                if cur.left:
                    q.append(cur.left)
                if cur.right:
                    q.append(cur.right)
            res.append(level)
        return res

        '''# 递归
        res = []
        
        def recur(node, depth):
            if not node:
                return
            if len(res) == depth:
                res.append([])
            res[depth].append(node.val)
            recur(node.left, depth+1)
            recur(node.right, depth+1)
        
        recur(root, 0)
        return res'''
# @lc code=end