这是我参与8月更文挑战的第25天，活动详情查看：8月更文挑战
描述
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.
Note:
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.
解析
根据题意，就是给出了一颗二叉树，和一个范围 [low, high] ，让我们求在这个范围内的节点的值的和是多少，思路比较简单，就是用递归的思想，判断每个节点的值如果在 [low, high] 中就累加起来，否则不去管他，最后递归的结果就是答案。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def rangeSumBST(self, root, low, high):
"""
:type root: TreeNode
:type low: int
:type high: int
:rtype: int
"""
if not root: return 0
def dfs(root):
if not root:
return 0
currentValue = root.val if low <= root.val <= high else 0
return currentValue+dfs(root.left)+dfs(root.right)
return dfs(root)
运行结果
Runtime: 296 ms, faster than 47.48% of Python online submissions for Range Sum of BST.
Memory Usage: 29.7 MB, less than 41.77% of Python online submissions for Range Sum of BST.
解析
上面的解法直接通过递归遍历了所有的节点，其实我们可以根据二叉树左子树小于根结点，右子树大雨根结点来减少计算量。仍然使用递归的思想。明显运行时间缩小了。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def rangeSumBST(self, root, low, high):
"""
:type root: TreeNode
:type low: int
:type high: int
:rtype: int
"""
if not root: return 0
if root.val>high:
return self.rangeSumBST(root.left, low, high)
if root.val<low:
return self.rangeSumBST(root.right, low, high)
return root.val + self.rangeSumBST(root.left, low, high) + self.rangeSumBST(root.right, low, high)
运行结果
Runtime: 260 ms, faster than 61.37% of Python online submissions for Range Sum of BST.
Memory Usage: 29.9 MB, less than 17.45% of Python online submissions for Range Sum of BST.
解析
同样的思路，不同的写法。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def rangeSumBST(self, root, low, high):
"""
:type root: TreeNode
:type low: int
:type high: int
:rtype: int
"""
def dfs(root):
if not root: return 0
if root.val>high:
return dfs(root.left)
if root.val<low:
return dfs(root.right)
return root.val + dfs(root.left) + dfs(root.right)
return dfs(root)
运行结果
Runtime: 248 ms, faster than 79.16% of Python online submissions for Range Sum of BST.
Memory Usage: 29.6 MB, less than 69.89% of Python online submissions for Range Sum of BST.
原题链接：leetcode.com/problems/ra…
您的支持是我最大的动力
