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:

1. The number of nodes in the tree is at most 10000.

2. The final answer is guaranteed to be less than 2^31.

解题要点：

设置一个全局变量sum，因为我们要在本function中进行递归，累计加符合要求的sum。然后检测当前val是否在这个range里，如果是，加进去；如不是，判断大小从而决定去树的左node还是右node。这是我个人的写法，专业写法的递归需要另外单写一个function。

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

        if(L <= root.val && root.val <= R){
            sum += root.val;
        }
        if(L < root.val) rangeSumBST(root.left, L, R);
        if(R > root.val) rangeSumBST(root.right, L, R);

        return sum;
    }

}