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107.py
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107.py
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# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# 层序遍历,头插法
def levelOrderBottom(self, root: 'TreeNode') -> 'List[List[int]]':
if root is None:return []
res = [[root.val]]
queue = []
queue.append(root)
while len(queue) != 0:
n = len(queue)
flag = False # 以防加入空level
level = []
while n > 0:
node = queue.pop(0)
n -= 1
if node.left is not None:
level.append(node.left.val)
queue.append(node.left)
flag = True
if node.right is not None:
level.append(node.right.val)
queue.append(node.right)
flag = True
if flag:
res.insert(0, level)
return res
# 递归解法
def levelOrderBottom2(self, root: 'TreeNode') -> 'List[List[int]]':
res = []
self.level_order(root, -1, res)
return res
def level_order(self, node: 'TreeNode', level: 'int', res: 'List[List[int]]'):
if node is None:return
# 超出递归的长度表明是新的一层,则新添加数组
if level < 0 and abs(level) > len(res):
res.insert(0, [])
# 可以理解成每个node都能对应到树的level
res[level].append(node.val)
if node.left is not None:
self.level_order(node.left, level-1, res)
if node.right is not None:
self.level_order(node.right, level-1, res)
if __name__ == '__main__':
s = Solution()
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(2)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.left = TreeNode(4)
root.right.right = TreeNode(3)
'''
输入:
1
/ \
2 2
/ \ / \
3 4 4 3
'''
print(s.levelOrderBottom2(root))