動機
就bfs
Problem
Given the root
of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).
Example 1:
Input: root = [3,9,20,null,null,15,7]Output: [[15,7],[9,20],[3]]
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
class Solution:
def levelOrderBottom(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)
ret.reverse()
return ret