"""
if s is None and t is None:
return True
elif s is None or t is None:
return False
return self.equals(s, t) or self.isSubtree(
s.left, t) or self.isSubtree(s.right, t)
def equals(self, s, t):
if s is None and t is None:
return True
elif s is None or t is None:
return False
return s.val == t.val and self.equals(s.left, t.left) and self.equals(
s.right, t.right)
from util import TreeNode
solution = Solution()
assert solution.isSubtree(TreeNode.of([3, 4, 5, 1, 2]), TreeNode.of([4, 1, 2]))
assert not solution.isSubtree(TreeNode.of([3, 4, 5, 1, 2, None, None, 0]),
TreeNode.of([4, 1, 2]))
assert solution.isSubtree(
TreeNode.of([
1, None, 1, None, 1, None, 1, None, 1, None, 1, None, 1, None, 1, None,
1, None, 1, None, 1, 2
]), TreeNode.of([1, None, 1, None, 1, None, 1, None, 1, None, 1, 2]))
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# https://leetcode.com/problems/merge-two-binary-trees/description/
from util import TreeNode
class Solution(object):
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if t1 is None and t2 is None:
return None
root = TreeNode(0)
if t1:
root.val += t1.val
if t2:
root.val += t2.val
root.left = self.mergeTrees(t1.left if t1 else None, t2.left if t2 else None)
root.right = self.mergeTrees(t1.right if t1 else None, t2.right if t2 else None)
return root
solution = Solution()
assert solution.mergeTrees(TreeNode.of([1,3,2,5]), TreeNode.of([2,1,3,None,4,None,7])) == [3,4,5,5,4,None,7]
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# https://leetcode.com/problems/diameter-of-binary-tree/description/
class Solution(object):
def diameterOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return 0
return self.inner(root)[0]-1
def inner(self, root):
"""return (max_diameter, max_depth)"""
if root is None:
return (0, 0)
l_diameter, l_depth = self.inner(root.left)
r_diameter, r_depth = self.inner(root.right)
return (max(l_diameter, r_diameter, l_depth+r_depth+1) , max(l_depth, r_depth)+1)
from util import TreeNode
solution = Solution()
assert solution.diameterOfBinaryTree(TreeNode.of([1,2,3,4,5])) == 3
# condition to optimize
from util import TreeNode
class Solution(object):
def pathSum(self, root, sum):
return self.inner(root, sum, 0, {0: 1})
def inner(self, root, target, sofar, counts):
if not root:
return 0
ret = 0
complement = sofar + root.val - target
if complement in counts:
ret += counts[complement]
sofar += root.val
counts.setdefault(sofar, 0) # consider root.val is 0, what happens?
counts[sofar] += 1
ret += self.inner(root.left, target, sofar, counts)
ret += self.inner(root.right, target, sofar, counts)
counts[sofar] -= 1
return ret
solution = Solution()
assert solution.pathSum(
TreeNode.of([10, 5, -3, 3, 2, None, 11, 3, -2, None, 1]), 8) == 3
# -*- coding: utf-8 -*-
# https://leetcode.com/problems/convert-bst-to-greater-tree/description/
from util import TreeNode
class Solution(object):
def convertBST(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
return self.rec(root, 0)[1]
def rec(self, root, sum):
if root is None:
return (sum, None)
total, r = self.rec(root.right, sum)
new_root = TreeNode(root.val + total)
new_total, l = self.rec(root.left, new_root.val)
new_root.left = l
new_root.right = r
return (new_total, new_root)
solution = Solution()
assert solution.convertBST(TreeNode.of([5, 2, 13])) == [18, 20, 13]
assert solution.convertBST(TreeNode.of([2, 0, 3, -4, 1])) == [5, 6, 3, 2, 6]
