题目描述：

给定二叉搜索树的根结点 root，返回值位于范围 [low, high] 之间的所有结点的值的和。

示例 1：

输入：root = [10,5,15,3,7,null,18], low = 7, high = 15
输出：32
示例 2：

输入：root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
输出：23

提示：

树中节点数目在范围 [1, 2 * 10^4] 内
1 <= Node.val <= 10^5
1 <= low <= high <= 10^5
所有 Node.val 互不相同

/**
 * 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 {
    List<Integer> list=new ArrayList<Integer>();
    public int rangeSumBST(TreeNode root, int low, int high) {
        inOrderTraverse(root);
        int sum=0;
        for(int i=0;i<list.size();i++)
        {
            int num=list.get(i);
            if(num>=low&&num<=high)
                sum+=num;
        }
        return sum;
    }

    public void inOrderTraverse(TreeNode root)
    {
        if(root==null)
            return;
        inOrderTraverse(root.left);
        list.add(root.val);
        inOrderTraverse(root.right);
    }

}

/**
 * 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 {
    public int rangeSumBST(TreeNode root, int low, int high) {
        if(root==null)
            return 0;
        if(root.val<low)
            return rangeSumBST(root.right,low,high);
        if(root.val>high)
            return rangeSumBST(root.left,low,high);
        return root.val+rangeSumBST(root.right,low,high)+rangeSumBST(root.left,low,high);
    }
}