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938. Range Sum of BST

Difficulty Topics

Description

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 * 104].
  • 1 <= Node.val <= 105
  • 1 <= low <= high <= 105
  • All Node.val are unique.

Solution

range-sum-of-bst.py
# 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 go(node):
            if not node: return 0

            return (node.val if low <= node.val <= high else 0) + go(node.left) + go(node.right)


        return go(root)
range-sum-of-bst.cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int rangeSumBST(TreeNode* root, int low, int high) {
        return helper(root, low, high);
    }

    int helper(TreeNode *node, int low, int high){
        if (node == nullptr) return 0;
        int left = helper(node->left, low, high);
        int right = helper(node->right, low, high);

        return ((node->val >= low && node->val <= high) ? node->val : 0 ) + left + right;
    }
};