[
tree
dfs
bfs
recursion
bst
]
BinarySearch 0134 Count BST Nodes in a Range
Problem statement
https://binarysearch.com/problems/Count-BST-Nodes-in-a-Range/
Solution
Variation of Leetcode 0938. Range Sum of BST.
Complexity
It is O(n)
for time and O(h)
for space.
Code
class Solution:
def solve(self, root, low, high):
def dfs(node):
if not node: return
if low <= node.val <= high: self.out += 1
if node.val > low: dfs(node.left)
if node.val < high: dfs(node.right)
self.out = 0
dfs(root)
return self.out