Given the root of a binary tree, return the number of uni-value subtrees (A subtree of treeName is a tree consisting of a node in treeName and all of its descendants).
A uni-value subtree means all nodes of the subtree have the same value.
Example 1:
5
/ \
1 (5)
/ \ \
(5) (5) (5)
Input: root = [5,1,5,5,5,null,5]
Output: 4
Example 2:
Input: root = []
Output: 0
Example 3:
Input: root = [5,5,5,5,5,null,5]
Output: 6
Constraints:
- The number of the node in the tree will be in the range
[0, 1000].
-1000 <= Node.val <= 1000
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# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def is_valid_part(self, root: Optional[TreeNode]) -> bool:
if not root:
return True
if root.left == root.right:
# leaf
self.total += 1
return True
result = True
if root.left:
left_all_same = self.is_valid_part(root.left)
if not left_all_same or root.val != root.left.val:
result = False
if root.right:
right_all_same = self.is_valid_part(root.right)
if not right_all_same or root.val != root.right.val:
result = False
if result:
self.total += 1
return result
def countUnivalSubtrees(self, root: Optional[TreeNode]) -> int:
self.total = 0
self.is_valid_part(root)
return self.total
'''
Shorter
'''
class Solution:
def is_valid_part(
self,
node: Optional[TreeNode],
parent_val: int
) -> bool:
if not node:
return True
is_left_valid = self.is_valid_part(node.left, node.val)
is_right_valid = self.is_valid_part(node.right, node.val)
# be careful of Short-Circuit Evaluation
if not is_left_valid or not is_right_valid:
return False
self.count += 1
return node.val == parent_val
def countUnivalSubtrees(self, root: Optional[TreeNode]) -> int:
self.count = 0
self.is_valid_part(root, 0)
return self.count
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