938. Range Sum of BST
DFS TREE
Given the root
node of a binary search tree and two integers low
and high
, return the sum of values of all nodes with a value in the inclusive range [low, high]
.
Example 1:
Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
Output: 32
Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.
Example 2:
Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
Output: 23
Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.
Constraints:
The number of nodes in the tree is in the range
[1, 2 * 10
4
]
.1 <= Node.val <= 10
5
1 <= low <= high <= 10
5
All
Node.val
are unique.
分析
判断根在不在范围内,在的话返回root+left+right, 否则按照范围返回左或者右
# 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 rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
def dfs(root):
if not root:
return 0
if low<=root.val <= high:
return root.val + dfs(root.left) + dfs(root.right)
elif root.val<low:
return dfs(root.right)
else:
return dfs(root.left)
return dfs(root)
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