Q938 Range Sum of BST

Range Sum of BST

Question

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:

1. The number of nodes in the tree is at most 10000.
2. The final answer is guaranteed to be less than 2^31.

Approach 1: Depth First Search

Intuition and Algorithm

We traverse the tree using a depth first search. If node.val falls outside the range [L, R], (for example node.val < L), then we know that only the right branch could have nodes with value inside [L, R].

We showcase two implementations - one using a recursive algorithm, and one using an iterative one.

Recursive Implementation

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rangeSumBST(self, root, L, R):
        def dfs(node):
            if node:
                if L <= node.val <= R:
                    self.ans += node.val
                if L < node.val:
                    dfs(node.left)
                if node.val < R:
                    dfs(node.right)

        self.ans = 0
        dfs(root)
        return self.ans

Iterative Implementation

class Solution(object):
    def rangeSumBST(self, root, L, R):
        ans = 0
        stack = [root]
        while stack:
            node = stack.pop()
            if node:
                if L <= node.val <= R:
                    ans += node.val
                if L < node.val:
                    stack.append(node.left)
                if node.val < R:
                    stack.append(node.right)
        return ans