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Time Complexity: O(n), where n is the number of nodes in the binary tree.
Space Complexity: O(h), where h is the height of the tree due to the recursion stack.
1class TreeNode:
2 def __init__(self, val=0, left=None, right=None):
3 self.val = val
4 self.left = left
5 self.right = right
6
7class Solution:
8 def dfs(self, node, maxVal):
9 if not node:
10 return 0
11 good = 1 if node.val >= maxVal else 0
12 maxVal = max(maxVal, node.val)
13 return good + self.dfs(node.left, maxVal) + self.dfs(node.right, maxVal)
14
15 def goodNodes(self, root):
16 return self.dfs(root, root.val)
17This Python solution utilizes a helper function dfs within the class Solution. It employs recursion to traverse the tree and count the good nodes as per the conditions provided.
Time Complexity: O(n), where n is the number of nodes in the tree.
Space Complexity: O(w), where w is the maximum width of the tree.
This JavaScript solution applies an iterative BFS approach using an array as the queue. Each object in the queue contains the current node and its max value, and during traversal, it checks and updates the count of good nodes.