1. 力扣938：二叉搜索树的范围和

1.1 题目：

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 * 104] 内
• 1 <= Node.val <= 105
• 1 <= low <= high <= 105
• 所有 Node.val 互不相同

1.2 思路：

没啥好说的，中序遍历加起来就好了。

1.3 题解：

/**
 * 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 {
    Deque<Integer> deque = new LinkedList<>();
    public int rangeSumBST(TreeNode root, int low, int high) {
        midTraverse(root, low, high);
        int sum = 0;
        while (!deque.isEmpty()){
            sum += deque.pop();
        }
        return sum;
    }
    private void midTraverse(TreeNode node, int low, int high) {
        if (node == null) {
            return;
        }
        midTraverse(node.left, low, high);
        if(node.val >= low && node.val <= high) {
            deque.push(node.val);
        }
        midTraverse(node.right, low, high);
    }
}