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102. Binary Tree Level Order Traversal ​

题目 ​

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
Related Topics
  • 树
  • 广度优先搜索
  • 二叉树

  • 👍 2155
  • 👎 0

    • 思路 ​

    通过对列进行层序遍历

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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 not root:
            return []

        res = []

        def tarverse(quene):
            if not quene:
                return
            size = len(quene)
            levels = []
            while size > 0:
                size -= 1
                node = quene.pop(0)
                levels.append(node.val)
                if node.left:
                    quene.append(node.left)
                if node.right:
                    quene.append(node.right)
            res.append(levels)

            tarverse(quene)

        tarverse([root])
        return res

# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)