Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
return its bottom-up level order traversal as:
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[
[15,7],
[9,20],
[3]
]
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- Use queue to do level traversal + deque.appendleft or list[ : : -1] to realize bottom-up
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from collections import deque
class :
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res = deque()
q = []
q.append(root)
while(q):
temp = []
l = len(q)
for _ in range(l):
cur = q.pop(0)
temp.append(cur.val)
if cur.left: q.append(cur.left)
if cur.right: q.append(cur.right)
res.appendleft(temp)
return res
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class :
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root: return []
res = []
q = []
q.append(root)
while(q):
temp = []
l = len(q)
for _ in range(l):
cur = q.pop(0)
temp.append(cur.val)
if cur.left: q.append(cur.left)
if cur.right: q.append(cur.right)
res.append(temp)
return res[::-1]
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Time complexity is O(n)
- DFS + record level
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class :
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
res = []
self.dfs(res, 0, root)
return res
def dfs(self, res, level, node):
if not node:
return
else:
if len(res) < (level + 1):
res.insert(0, [])
res[-(level+1)].append(node.val)
self.dfs(res, level + 1, node.left)
self.dfs(res, level + 1, node.right)
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So time complexity is O(n^2)
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