标签:right val BST Sum high Range low root self
Given the root
node of a binary search tree, return the sum of values of all nodes with a value in the range [low, high]
.
Example 1:
Input: root = [10,5,15,3,7,null,18], low = 7, high = 15 Output: 32
Example 2:
Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10 Output: 23
Constraints:
- The number of nodes in the tree is in the range
[1, 2 * 104]
. 1 <= Node.val <= 105
1 <= low <= high <= 105
- All
Node.val
are unique.
# 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: TreeNode, low: int, high: int) -> int: if not root: return 0 self.values = 0 self.dfs(root, low, high) return self.values def dfs(self, root, low, high): if not root: return if low <= root.val <= high: self.values+= root.val self.dfs(root.left, low, high) self.dfs(root.right, low, high)
方法二:利用了BST特性的
# 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: TreeNode, low: int, high: int) -> int: if not root: return 0 if root.val > high: return self.rangeSumBST(root.left, low, high) elif root.val < low: return self.rangeSumBST(root.right, low, high) else: return root.val + self.rangeSumBST(root.left, low, high) + self.rangeSumBST(root.right, low, high)
标签:right,val,BST,Sum,high,Range,low,root,self 来源: https://www.cnblogs.com/sky37/p/13983270.html
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