題目，這一題會給 low 和 high，只要這棵樹的節點數值滿足大於等於 low 和小於等於 high 就將其相加並成為最後結果。思路是使用前序進行遍歷，並利用 BST 特性左小右大來決定下個節點是否要進行遞歸。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int res = 0;
public int rangeSumBST(TreeNode root, int low, int high) {
if (root != null){
if (root.val >= low && root.val <= high){
res += root.val;
}
if (root.val >= low){
rangeSumBST(root.left, low, high);
}
if (root.val <= high){
rangeSumBST(root.right, low, high);
}
}
return res;
}
}