LeetCode 938: Range Sum of BST
LeetCode 938 Solution Explanation
Explanation
To solve this problem, we can perform a depth-first search (DFS) traversal of the binary search tree (BST). At each node, we check if the node's value falls within the range [low, high]. If it does, we add the node's value to the sum. We then recursively traverse the left and right subtrees if they exist.
Algorithm:
- Initialize a variable
sumto store the total sum. - Perform a recursive DFS traversal starting from the root node.
- At each node:
- If the node's value is within the range [low, high], add the value to the
sum. - Recursively traverse the left and right subtrees if they exist.
- If the node's value is within the range [low, high], add the value to the
- Return the final
sumvalue.
Time Complexity:
The time complexity of this algorithm is O(N), where N is the number of nodes in the binary search tree. We visit each node once in the worst case.
Space Complexity:
The space complexity is O(H), where H is the height of the binary search tree. In the worst case, the recursion stack can go as deep as the height of the tree.
LeetCode 938 Solutions in Java, C++, Python
class Solution {
public int rangeSumBST(TreeNode root, int low, int high) {
if (root == null) {
return 0;
}
int sum = 0;
if (root.val >= low && root.val <= high) {
sum += root.val;
}
sum += rangeSumBST(root.left, low, high);
sum += rangeSumBST(root.right, low, high);
return sum;
}
}Interactive Code Editor for LeetCode 938
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