LeetCode 938 - Range Sum of BST (Easy)

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.

 方法：一开始没有仔细审题，没有用到BST的特性，用了最straightforward的方法来做dfs。

time complexity: O(N)  space  complexity :O(N) 

# 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)