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kth-smallest-element-in-a-bst solution
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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 kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
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# DFS
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# 시간복잡도 최악 O(n)/ 공간복잡도 O(log n) ~ O(n) (재귀깊이)
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self.count = 0
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self.answer = None
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def dfs(node):
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if not node:
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return
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# 가장 작은 값부터(왼쪽서브트리) 재귀로 탐색
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dfs(node.left)
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self.count += 1
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# k번째 작은 값 찾으면 바로 종료하기
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if self.count == k:
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self.answer = node.val
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return
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# 오른쪽 서브트리 탐색
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dfs(node.right)
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dfs(root)
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return self.answer

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