forked from shuboc/LeetCode-2
-
Notifications
You must be signed in to change notification settings - Fork 1
/
binary-tree-pruning.py
53 lines (49 loc) · 1.32 KB
/
binary-tree-pruning.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
# Time: O(n)
# Space: O(h)
# We are given the head node root of a binary tree,
# where additionally every node's value is either a 0 or a 1.
#
# Return the same tree where every subtree (of the given tree)
# not containing a 1 has been removed.
#
# (Recall that the subtree of a node X is X,
# plus every node that is a descendant of X.)
#
# Example 1:
# Input: [1,null,0,0,1]
# Output: [1,null,0,null,1]
#
# Explanation:
# Only the red nodes satisfy the property "every subtree not containing a 1".
# The diagram on the right represents the answer.
#
# Example 2:
# Input: [1,0,1,0,0,0,1]
# Output: [1,null,1,null,1]
#
# Example 3:
# Input: [1,1,0,1,1,0,1,0]
# Output: [1,1,0,1,1,null,1]
#
# Note:
# - The binary tree will have at most 100 nodes.
# - The value of each node will only be 0 or 1.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pruneTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return None
root.left = self.pruneTree(root.left)
root.right = self.pruneTree(root.right)
if not root.left and not root.right and root.val == 0:
return None
return root