38 lines
1.3 KiB
Python
38 lines
1.3 KiB
Python
# Time: O(N)
|
|
# Space: O(N)
|
|
|
|
# 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 goodNodes(self, root: TreeNode) -> int:
|
|
def dfs(node, max_value) -> int:
|
|
'''
|
|
Perform pre-order traversal and keep track of max
|
|
elements in the tree. Any subsequent traversal can
|
|
then compare against the updated max_value to see if
|
|
it's a good node or not
|
|
'''
|
|
|
|
# If no left/right nodes, then we can just return 0
|
|
if not node: return 0
|
|
|
|
# Current node is good if it's value is greater than or
|
|
# equal to the `max_value` seen so far
|
|
res = 1 if max_value <= node.val else 0
|
|
|
|
# Compute the new max value, the current node could be it
|
|
max_value = max(max_value, node.val)
|
|
|
|
# Do traversal on left and right nodes and add their units
|
|
res += dfs(node.left, max_value) + dfs(node.right, max_value)
|
|
|
|
# This will indicate the count
|
|
return res
|
|
|
|
|
|
return dfs(root, root.val)
|