38 lines
1.3 KiB
Python
38 lines
1.3 KiB
Python
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# Time: O(N)
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# Space: O(N)
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def goodNodes(self, root: TreeNode) -> int:
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def dfs(node, max_value) -> int:
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'''
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Perform pre-order traversal and keep track of max
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elements in the tree. Any subsequent traversal can
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then compare against the updated max_value to see if
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it's a good node or not
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'''
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# If no left/right nodes, then we can just return 0
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if not node: return 0
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# Current node is good if it's value is greater than or
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# equal to the `max_value` seen so far
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res = 1 if max_value <= node.val else 0
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# Compute the new max value, the current node could be it
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max_value = max(max_value, node.val)
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# Do traversal on left and right nodes and add their units
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res += dfs(node.left, max_value) + dfs(node.right, max_value)
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# This will indicate the count
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return res
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return dfs(root, root.val)
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