動機

複習bfs

Problem

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

Sol

以level為單位，當成state 這樣在解103會有好處

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        q = deque([[root]])
        ret = []
        while q:
            now = []
            rs = q.popleft()
            ret.append([r.val for r in rs])
            for r in rs:
                for n in [r.left,r.right]:
                    if n:
                        now.append(n)
            if len(now) > 0:
                q.append(now)
        return ret