【LeetCode】938. Range Sum of BST 解题报告（Python)
题目描述
Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).
The binary search tree is guaranteed to have unique values.
Example 1:
Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32
Example 2:
Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23
Note:
The number of nodes in the tree is at most 10000.
The final answer is guaranteed to be less than 2^31.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def rangeSumBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: int
"""
if not root:
return 0
res = 0
if L <= root.val <= R:
res += root.val
res += self.rangeSumBST(root.left, L, R)
res += self.rangeSumBST(root.right, L, R)
elif root.val < L:
res += self.rangeSumBST(root.right, L, R)
elif root.val > R:
res += self.rangeSumBST(root.left, L, R)
return res
简化代码：直接判断寻找的方向。如果root节点小于R，说明右边可以继续搜索；如果root节点大于L，说明左边可以继续搜索。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def rangeSumBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: int
"""
res = [0]
self.dfs(root, L, R, res)
return res[0]
def dfs(self, root, L, R, res):
if not root:
return
if L <= root.val <= R:
res[0] += root.val
if root.val < R:
self.dfs(root.right, L, R, res)
if root.val > L:
self.dfs(root.left, L, R, res)
参考来源：https://blog.csdn.net/fuxuemingzhu/article/details/83961263