938. Range Sum of BST

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:

Generally for a tree problem, we can think of either using a recursive traversal, or bfs.

We choose to use recursion here.

Knowing the property of BST and given a root node.
1. The exit condition is that when root is null, we return 0.
2. If root.val < L, we know all nodes in its left subtree are smaller than L. So we do not need to count that. We only consider its right subtree.
3. If root.val > R, we know all nodes in its right subtree are larger than R. So we do not need to count that. We only consider its left subtree.
4. If root.val is between L and R, we need to count this value. And we need to count both its left and right subtrees recursively.

The time complexity is O(n), assume n is the total number of nodes, since each node will be traversed only once.


Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int rangeSumBST(TreeNode root, int L, int R) {
        if (root == null) return 0;
        if (root.val < L) return rangeSumBST(root.right, L, R);
        if (root.val > R) return rangeSumBST(root.left, L, R);
        return root.val 
            + rangeSumBST(root.right, L, R) 
            + rangeSumBST(root.left, L, R);
    }
}