938. Range Sum of BST
Tags: ‘Tree’, ‘Recursion’
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:
- The number of nodes in the tree is at most
10000. - The final answer is guaranteed to be less than
2^31.
Solution
//Recursive
public int rangeSumBST(TreeNode root, int L, int R) {
if (root == null) return 0;
int sum = 0;
if (root.val >= L && root.val <= R) {
sum += root.val; //preorder
}
if (root.val > L) {
sum += rangeSumBST(root.left, L, R);
}
if (root.val < R) {
sum += rangeSumBST(root.right, L, R);
}
return sum;
}
// Iterative
public int rangeSumBST(TreeNode root, int L, int R) {
if (root == null) return 0;
int sum = 0;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node == null) continue;
if (node.val >= L && node.val <= R) {
sum += node.val;
}
if (node.val > L) {
// left is possible
stack.push(node.left);
}
if (node.val < R) {
// right is possible
stack.push(node.right);
}
}
return sum;
}