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。

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

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

代码如下:

/** * 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 static int rangeSumBST(TreeNode root, int low, int high) { 
        ArrayList<Integer> list = new ArrayList<>();
        mid(root, list);
        int ans = 0;
        for (Integer k : list) { 
            if (k >= low && k <= high) { 
                ans += k;
            }
        }
        return ans;
    }

    public static void mid(TreeNode root, ArrayList<Integer> list) { 
        if (root == null) { 
            return;
        }
        mid(root.left, list);
        list.add(root.val);
        mid(root.right, list);
    }
}

执行结果:

本文地址:https://blog.csdn.net/FYPPPP/article/details/114272445